From b5fbed36531d012d8b2c316795af98334da10566 Mon Sep 17 00:00:00 2001 From: Ruvikm <1092045998@qq.com> Date: Wed, 10 Feb 2021 12:03:45 +0800 Subject: [PATCH] =?UTF-8?q?=E6=95=B4=E7=90=86=E4=BA=86=E7=BA=BF=E6=80=A7?= =?UTF-8?q?=E8=A1=A8=E9=A1=BA=E5=BA=8F=E5=AD=98=E5=82=A8=E7=9A=84=E4=BB=A3?= =?UTF-8?q?=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 线性表/顺序实现大题/P19.1.cpp | 82 +++++ 线性表/顺序实现大题/P19.3.cpp | 71 +++++ 线性表/顺序实现大题/P19.4.cpp | 76 +++++ 线性表/顺序实现大题/P19.5.cpp | 76 +++++ 线性表/顺序实现大题/P19.6.cpp | 72 +++++ 线性表/顺序实现大题/P19.7.cpp | 85 +++++ 线性表/顺序实现大题/P19.8.cpp | 65 ++++ 线性表/顺序实现大题/P20.10.cpp | 70 +++++ 线性表/顺序实现大题/P20.11.cpp | 84 +++++ 线性表/顺序实现大题/P20.12.cpp | 80 +++++ 线性表/顺序实现大题/P20.9.cpp | 118 +++++++ 线性表/顺序实现大题/P21.13.cpp | 76 +++++ 线性表/顺序实现大题/p19.2.cpp | 66 ++++ 线性表/顺序实现大题/线性表顺序表示大题.cpp | 343 --------------------- 14 files changed, 1021 insertions(+), 343 deletions(-) create mode 100644 线性表/顺序实现大题/P19.1.cpp create mode 100644 线性表/顺序实现大题/P19.3.cpp create mode 100644 线性表/顺序实现大题/P19.4.cpp create mode 100644 线性表/顺序实现大题/P19.5.cpp create mode 100644 线性表/顺序实现大题/P19.6.cpp create mode 100644 线性表/顺序实现大题/P19.7.cpp create mode 100644 线性表/顺序实现大题/P19.8.cpp create mode 100644 线性表/顺序实现大题/P20.10.cpp create mode 100644 线性表/顺序实现大题/P20.11.cpp create mode 100644 线性表/顺序实现大题/P20.12.cpp create mode 100644 线性表/顺序实现大题/P20.9.cpp create mode 100644 线性表/顺序实现大题/P21.13.cpp create mode 100644 线性表/顺序实现大题/p19.2.cpp delete mode 100644 线性表/顺序实现大题/线性表顺序表示大题.cpp diff --git a/线性表/顺序实现大题/P19.1.cpp b/线性表/顺序实现大题/P19.1.cpp new file mode 100644 index 0000000..b6b8742 --- /dev/null +++ b/线性表/顺序实现大题/P19.1.cpp @@ -0,0 +1,82 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int* a, int* b) { + int t; + t = *a; + *a = *b; + *b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List* list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(&list->data[start + i], &list->data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.1 +//从顺序表中删除具有最小值的元素(假设唯一)并由函数返回被删元素的值。空出的位 +//置由最后一个元素填补, 若顺序表为空则显示出错信息并退出运行 + +int FindMin(List& list) +{ + if (!list.length) { + printf("Error!\n"); + return -1; + } + + int Min = 9999; + int point = 0; + for (int i = 0; i < list.length; i++) + { + if (list.data[i] < Min) + { + Min = list.data[i]; + point = i; + } + } + list.data[point] = list.data[list.length - 1]; + list.length -= 1; + return Min; +} + +int main() +{ + List list; + list.length = 4; + int Data[] = { 2,4,3,9 }; + for (int i = 0; i < list.length; i++) + list.data[i] = Data[i]; + PrintList(list); + int del = 2; + cout <<"删除最小值的元素为："<< FindMin(list) << endl; + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P19.3.cpp b/线性表/顺序实现大题/P19.3.cpp new file mode 100644 index 0000000..c457b02 --- /dev/null +++ b/线性表/顺序实现大题/P19.3.cpp @@ -0,0 +1,71 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.3 +//对长度为n的顺序表L, 编写一个时间复杂度为O(m)空间复杂度为O(1)的算法, +//该算法删除线性表中所有值为x的数据元素 + +void DeleteX(List& list, int X) { + int k = 0, i = 0; + while (i < list.length) { + if (list.data[i] == X) + k++; + else + list.data[i - k] = list.data[i]; + i++; + } + list.length = list.length - k; +} + +int main() +{ + List list; + list.length = 9; + int Data[] = { 2,4,3,9,3,12,3,76,3 }; + for (int i = 0; i < list.length; i++) + list.data[i] = Data[i]; + PrintList(list); + DeleteX(list, 3); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P19.4.cpp b/线性表/顺序实现大题/P19.4.cpp new file mode 100644 index 0000000..f7bf9e1 --- /dev/null +++ b/线性表/顺序实现大题/P19.4.cpp @@ -0,0 +1,76 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.4 +//从有序顺序表中删除其值在给定值s与t之间(要求s < t)的所有元素, 如果s或t不合 +//理或顺序表为空, 则显示出错信息并退出运行 + +void OrderlyDelete(List& list, int X, int T) { + int i = 0, j = 0; + for (i; i < list.length; i++) + if (list.data[i] >= X) + break; + + for (j; j < list.length; j++) { + if (list.data[j] > T) + break; + } + int len = j - 1; + for (i; i < list.length; i++, j++) { + list.data[i] = list.data[j]; + } + list.length -= len; +} + +int main() +{ + List list; + list.length = 9; + int Data[] = { 2,4,6,9,11,12,24,34,44 }; + for (int i = 0; i < list.length; i++) + list.data[i] = Data[i]; + PrintList(list); + OrderlyDelete(list, 4, 24); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P19.5.cpp b/线性表/顺序实现大题/P19.5.cpp new file mode 100644 index 0000000..dd30ba1 --- /dev/null +++ b/线性表/顺序实现大题/P19.5.cpp @@ -0,0 +1,76 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.5 +//从顺序表中删除其值在给定值s与t之间(包含s和t, 要求s < t)的所有元素, +//如果s或t不合理或顺序表为空, 则显示出错信息并退出运行 + +void DeleteS_T(List& list, int S, int T) { + if (S >= T || !list.length) + return; + + for (int i = 0; i < list.length; i++) { + if (list.data[i] >= S && list.data[i] <= T) { + for (int j = i + 1; j < list.length; j++) { + list.data[j - 1] = list.data[j]; + } + //要重新回退一个位置，重新检测，否则会跳过某些数字 + if (i) + i -= 1; + list.length -= 1; + } + } +} + +int main() +{ + List list; + list.length = 9; + int Data[] = { 2,4,9,5,11,5,24,14,44 }; + for (int i = 0; i < list.length; i++) + list.data[i] = Data[i]; + PrintList(list); + DeleteS_T(list, 4, 11); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P19.6.cpp b/线性表/顺序实现大题/P19.6.cpp new file mode 100644 index 0000000..658128d --- /dev/null +++ b/线性表/顺序实现大题/P19.6.cpp @@ -0,0 +1,72 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.6 +//从有序顺序表中删除所有其值重复的元素,使表中所有元素的值均不同 + +void OlderlyDeleteSame(List& list) { + + _for(i, 0, list.length) { + if (list.data[i + 1] == list.data[i]) { + for (int j = i + 1; j < list.length; j++) { + list.data[j] = list.data[j + 1]; + } + list.length -= 1; + if (i) + i -= 1; + } + } +} + +int main() +{ + List list; + list.length = 9; + int Data[] = { 1,4,9,9,9,9,9,24,14,3 }; + for (int i = 0; i < list.length; i++) + list.data[i] = Data[i]; + PrintList(list); + OlderlyDeleteSame(list); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P19.7.cpp b/线性表/顺序实现大题/P19.7.cpp new file mode 100644 index 0000000..9896d65 --- /dev/null +++ b/线性表/顺序实现大题/P19.7.cpp @@ -0,0 +1,85 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.7 +//将两个有序顺序表合并为一个新的有序顺序表, 并由函数返回结果顺序表 + +List MergeList(List A, List B) { + int i = 0, j = 0; + List C; + int k = 0; + while (i < A.length && j < B.length) { + if (A.data[i] >= B.data[j]) { + C.data[k++] = B.data[j++]; + } + else { + C.data[k++] = A.data[i++]; + } + } + while (i < A.length) { + C.data[k++] = A.data[i++]; + } + while (j < B.length) { + C.data[k++] = B.data[j++]; + } + C.length = k; + return C; +} + +int main() +{ + List list1, list2; + list1.length = 9; + int Data1[] = { 1,4,9,11,14,24,39,54,76,99 }; + _for(i, 0, list1.length) + list1.data[i] = Data1[i]; + PrintList(list1); + + list2.length = 5; + int Data2[] = { 11,24,43,65,71 }; + _for(i, 0, list2.length) + list2.data[i] = Data2[i]; + PrintList(list2); + PrintList(MergeList(list1, list2)); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P19.8.cpp b/线性表/顺序实现大题/P19.8.cpp new file mode 100644 index 0000000..82c4171 --- /dev/null +++ b/线性表/顺序实现大题/P19.8.cpp @@ -0,0 +1,65 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.8 +//已知在一维数组A[m + n]中依次存放两个线性表(a1, a2, a3…, am)和(b, b2, b3, …, bn)。试编 +//写一个函数, 将数组中两个顺序表的位置互换, 即将(b1,b2,b3,…, bn)放在(a1，a2, a3, …, am)的前面 + +void swaplist(List& A, int m, int n, int size) { + Reserve(A, 0, m + n - 1, size); + Reserve(A, 0, n - 1, size); + Reserve(A, n, m + n - 1, size); +} + +int main() +{ + List list; + list.length = 10; + int Data[] = { 1,4,9,7,14,24,12,54,42,32 }; + _for(i, 0, list.length) + list.data[i] = Data[i]; + PrintList(list); + swaplist(list, 4, 6, list.length); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P20.10.cpp b/线性表/顺序实现大题/P20.10.cpp new file mode 100644 index 0000000..6a23991 --- /dev/null +++ b/线性表/顺序实现大题/P20.10.cpp @@ -0,0 +1,70 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P20.10 +//〖2010统考真题〗设将n(n > 1)个整数存放到一维数组R中。设计一个在时间和空间两 +//方面都尽可能高效的算法。将R中保存的序列循环左移p(0 +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P20.11 +//〖2011统考真题】一个长度为L(L≥1)的升序序列S, 处在第[L/2]个位置的数称为S的 +//中位数。例如, 若序列S1 = (1, 13, 15, 17, 19), 则S1的中位数是15, 两个序列的中位数 +//是含它们所有元素的升序序列的中位数。例如, 若S2 = (2, 4, 6, 8, 20), 则S1和S2的中位 +//数是11。现在有两个等长升序序列A和B, 试设计一个在时间和空间两方面都尽可能 +//高效的算法, 找出两个序列A和B的中位数。要求 +//1)给出算法的基本设计思想 +//2)根据设计思想, 采用C或C++或Java语言描述算法, 关键之处给出注释 +//3)说明你所设计算法的时间复杂度和空间复杂度 + +//改编的归并排序 +int FindMid(List A, List B) { + int i = 0, j = 0; + while (i + j != (A.length + B.length) / 2 - 1) { + if (A.data[i] <= B.data[j]) { + i++; + } + else { + j++; + } + } + return A.data[i] < B.data[j] ? A.data[i] : B.data[j]; +} + +int main() +{ + List list1, list2; + list1.length = 9; + int Data1[] = { 1,4,9,11,14,24,39,54,76,99 }; + _for(i, 0, list1.length) + list1.data[i] = Data1[i]; + PrintList(list1); + + list2.length = 5; + int Data2[] = { 11,24,43,65,71 }; + _for(i, 0, list2.length) + list2.data[i] = Data2[i]; + PrintList(list2); + cout << "list1与list2的中位数为：" << FindMid(list1, list2) << endl; +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P20.12.cpp b/线性表/顺序实现大题/P20.12.cpp new file mode 100644 index 0000000..966a102 --- /dev/null +++ b/线性表/顺序实现大题/P20.12.cpp @@ -0,0 +1,80 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P20.12 +//〖2013统考真题〗已知一个整数序列A = (a0, a1…, an-1), 其中0 ≤ ai < n(0 ≤i < n)。若 +// 存在 ap1 = ap2 = ... = apm = x 且 m > n / 2 (0≤ Pk < n,1≤ k ≤ m), 则称x为A的主元素。 +// 例如A=(0,5,5,3,5,7,5,5)，则5为主元素; 又如A = (0, 5, 5, 3, 5, 1, 5, 7), 则A中没有主 +// 元素。假设A中的n个元素保存在一个一维数组中, 请设计一个尽可能高效的算法, 找 +// 出A的主元素。若存在主元素, 则输岀该元素; 否则输出 - 1。要求: +//1)给出算法的基本设计思想 +//2)根据设计思想, 采用C或C++或Java语言描述算法, 关键之处给出注释 +//3)说明你所设计算法的时间复杂度和空间复杂度 + +//次优算法，用空间换时间，类似桶排 +int FindMain(List A, int n) { + int m_num = 0; + int* Box = new int[n](); //动态申请一个大小为n的数组 + _for(i, 0, A.length) { + Box[A.data[i]]++; + } + _for(i, 0, n) { + if (Box[i] > n / 2) { + return i; + } + } + return -1; +} + +int main() +{ + List list; + list.length = 10; + /*int Data[] = { 1,4,9,11,14,24,32,54,62,81 };*/ + int Data[] = { 1,3,2,6,3,3,3,3,3,1 }; + _for(i, 0, list.length) + list.data[i] = Data[i]; + PrintList(list); + cout << "list中的主元素为：" << FindMain(list, 10) << endl; +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P20.9.cpp b/线性表/顺序实现大题/P20.9.cpp new file mode 100644 index 0000000..643fd0d --- /dev/null +++ b/线性表/顺序实现大题/P20.9.cpp @@ -0,0 +1,118 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P20.9 +//线性表(a1,a2,a3,..,an)中的元素递增有序且按顺序存储于计算机内。 +//要求设计一算法, 完成用最少时间在表中查找数值为x的元素, 若找到则将其与后继元素位置相交换, +//若找不到则将其插入表中并使表中元素仍递增有序 + +//我的方法 大力出奇迹 +void CheckAndSwap(List& A, int x) { + bool flag = true; + _for(i, 0, A.length) { + if (A.data[i] == x) { + swap(A.data[i], A.data[i + 1]); + printf("Find it\n"); + flag = false; + break; + } + } + if (flag) { + printf("Not find\n"); + int i = 0; + for (i = 0; i < A.length && A.data[i] <= x; i++); + if (i < A.length) + { + for (int j = A.length; j >= i; j--) { + A.data[j + 1] = A.data[j]; + } + } + A.data[i] = x; + A.length += 1; + } +} + +//答案方法 折半查找 +void SearchExchangeInsert(List& A, int x) { + int low = 0, mid = A.length / 2, high = A.length - 1; + bool Isfind = false; + while (low <= high) { + mid = (low + high) / 2; + if (x == A.data[mid]) { + Isfind = true; + printf("Find\n"); + break; + } + if (x > A.data[mid]) { + low = mid + 1; + } + if (x < A.data[mid]) { + high = mid - 1; + } + } + if (Isfind && mid != A.length - 1) { + swap(A.data[mid], A.data[mid + 1]); + } + if (!Isfind) { + int i = 0; + for (i = A.length - 1; i > high; i--) { + A.data[i + 1] = A.data[i]; + } + A.data[i + 1] = x; + A.length += 1; + } +} + +int main() +{ + List list; + list.length = 10; + int Data[] = { 1,4,9,11,14,24,32,54,62,81 }; + _for(i, 0, list.length) + list.data[i] = Data[i]; + PrintList(list); + //CheckAndSwap(list, 54); + SearchExchangeInsert(list, 23); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/P21.13.cpp b/线性表/顺序实现大题/P21.13.cpp new file mode 100644 index 0000000..1e348ae --- /dev/null +++ b/线性表/顺序实现大题/P21.13.cpp @@ -0,0 +1,76 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P21.13 +//〖2018统考真题】给定一个含n(n≥1)个整数的数组, 请设计一个在时间上尽可能高效 +//的算法, 找出数组中未出现的最小正整数。例如, 数组{ -5,3,2,3 }中未出现的最小正整数 +//是1; 数组{ 1,2,3 }中未出现的最小正整数是4。要求 +//1)给出算法的基本设计思想 +//2)根据设计思想, 采用C或C++语言描述算法, 关键之处给出注释 +//3)说明你所设计算法的时间复杂度和空间复杂度 + +//用空间换时间 +int FindMinNum(List A, int n) { + int* Box = new int[n + 1](); + _for(i, 0, A.length) { + if (A.data[i] > 0 && A.data[i] <= n) + Box[A.data[i]]++; + } + _for(i, 1, n) { + if (!Box[i]) + return i; + } + return -1; +} + +int main() +{ + List list; + list.length = 10; + int Data[] = { 1,4,9,-11,14,-24,32,54,-62,81 }; + _for(i, 0, list.length) + list.data[i] = Data[i]; + PrintList(list); + cout << "list中未出现的最小正整数是为：" << FindMinNum(list, 10) << endl; +} \ No newline at end of file diff --git a/线性表/顺序实现大题/p19.2.cpp b/线性表/顺序实现大题/p19.2.cpp new file mode 100644 index 0000000..ceeb601 --- /dev/null +++ b/线性表/顺序实现大题/p19.2.cpp @@ -0,0 +1,66 @@ +#include +#include +#include + +using namespace std; + +#pragma region 建立顺序存储的线性表 + +#define MAX 30 +#define _for(i,a,b) for( int i=(a); i<(b); ++i) +#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) + +typedef struct { + int data[MAX]; + int length; +}List; + +void swap(int& a, int& b) { + int t; + t = a; + a = b; + b = t; +} + +//给定一个List，传入List的大小，要逆转的起始位置 +void Reserve(List& list, int start, int end, int size) { + if (end <= start || end >= size) { + return; + } + int mid = (start + end) / 2; + _rep(i, 0, mid - start) { + swap(list.data[start + i], list.data[end - i]); + } +} + +void PrintList(List list) { + _for(i, 0, list.length) { + cout << list.data[i] << " "; + } + cout << endl; +} + +#pragma endregion + +//P19.2 +//设计一个高效算法，将顺序表的所有元素逆置，要求算法的空间复杂度为O⑴ + +void reverse(List& list) { + if (!list.length) + return; + for (int i = 0; i < list.length / 2; i++) { + swap(list.data[i], (list.data[list.length - i - 1])); + } +} + +int main() +{ + List list; + list.length = 9; + int Data[] = { 2,4,3,9,43,12,54,76,38 }; + for (int i = 0; i < list.length; i++) + list.data[i] = Data[i]; + PrintList(list); + reverse(list); + PrintList(list); +} \ No newline at end of file diff --git a/线性表/顺序实现大题/线性表顺序表示大题.cpp b/线性表/顺序实现大题/线性表顺序表示大题.cpp deleted file mode 100644 index 1ffbe18..0000000 --- a/线性表/顺序实现大题/线性表顺序表示大题.cpp +++ /dev/null @@ -1,343 +0,0 @@ -#include -#include -#include - -using namespace std; - -#define MAX 30 -#define _for(i,a,b) for( int i=(a); i<(b); ++i) -#define _rep(i,a,b) for( int i=(a); i<=(b); ++i) - -typedef struct { - int data[MAX]; - int length; -}List; - -void swap(int* a, int* b) { - int t; - t = *a; - *a = *b; - *b = t; -} - -//给定一个List，传入List的大小，要逆转的起始位置 -void Reserve(List* list, int start, int end, int size) { - if (end <= start || end >= size) { - return; - } - int mid = (start + end) / 2; - _rep(i, 0, mid - start) { - swap(&list->data[start + i], &list->data[end - i]); - } - /* _for(i, 0, list->length) { - printf("%d ", list->data[i]); - }*/ -} - -//P19.1 -//从顺序表中删除具有最小值的元素(假设唯一)并由函数返回被删元素的值。空出的位 -//置由最后一个元素填补, 若顺序表为空则显示出错信息并退出运行 -int FindMin(List* list) -{ - if (!(*list).length) { - printf("Error!\n"); - return -1; - } - - int Min = 9999; - int point; - for (int i = 0; i < (*list).length; i++) - { - if ((*list).data[i] < Min) - { - Min = (*list).data[i]; - point = i; - } - } - (*list).data[point] = (*list).data[(*list).length - 1]; - (*list).length -= 1; - return Min; -} - -//P19.2 -//设计一个高效算法，将顺序表的所有元素逆置，要求算法的空间复杂度为O⑴ -void reverse(List* list) { - if (!list->length) - return; - for (int i = 0; i < (*list).length / 2; i++) { - swap(&(list->data[i]), &(list->data[list->length - i - 1])); - } -} - -//P19.3 -//对长度为n的顺序表L, 编写一个时间复杂度为O(m)空间复杂度为O(1)的算法, -//该算法删除线性表中所有值为x的数据元素 -void DeleteX(List* list, int X) { - for (int i = 0; i < (*list).length; i++) { - if ((*list).data[i] == X) { - for (int j = i + 1; j < (*list).length; j++) { - (*list).data[j - 1] = (*list).data[j]; - } - (*list).length -= 1; - break; - } - } -} - -//P19.4 -//从有序顺序表中删除其值在给定值s与t之间(要求s < t)的所有元素, 如果s或t不合 -//理或顺序表为空, 则显示出错信息并退出运行 -void OrderlyDelete(List* list, int X, int T) { - int i = 0, j = 0; - for (i; i < (*list).length; i++) - if ((*list).data[i] >= X) - break; - - for (j; j < (*list).length; j++) { - if ((*list).data[j] > T) - break; - } - int len = j - 1; - for (i; i < (*list).length; i++, j++) { - list->data[i] = list->data[j]; - } - list->length -= len; -} - -//P19.5 -//从顺序表中删除其值在给定值s与t之间(包含s和t, 要求s < t)的所有元素, -//如果s或t不合理或顺序表为空, 则显示出错信息并退出运行 -void DeleteS_T(List* list, int S, int T) { - if (S >= T || !list->length) - return; - - for (int i = 0; i < (*list).length; i++) { - if ((*list).data[i] >= S && (*list).data[i] <= T) { - for (int j = i + 1; j < (*list).length; j++) { - (*list).data[j - 1] = (*list).data[j]; - } - //要重新回退一个位置，重新检测，否则会跳过某些数字 - if (i) - i -= 1; - (*list).length -= 1; - } - } -} - -//P19.6 -//从有序顺序表中删除所有其值重复的元素,使表中所有元素的值均不同 - -void OlderlyDeleteSame(List* list) { - for (int i = 0; i < list->length; i++) { - if (list->data[i + 1] == list->data[i]) { - for (int j = i + 1; j < list->length; j++) { - list->data[j] = list->data[j + 1]; - } - list->length -= 1; - if (i) - i -= 1; - } - } -} - -//P19.7 -//将两个有序顺序表合并为一个新的有序顺序表, 并由函数返回结果顺序表 - -List MergeList(List A, List B) { - int i = 0, j = 0; - List C; - int k = 0; - while (i < A.length && j < B.length) { - if (A.data[i] >= B.data[j]) { - C.data[k++] = B.data[j++]; - } - else { - C.data[k++] = A.data[i++]; - } - } - while (i < A.length) { - C.data[k++] = A.data[i++]; - } - while (j < B.length) { - C.data[k++] = B.data[j++]; - } - C.length = k; - return C; -} - -//P19.8 -//已知在一维数组A[m + n]中依次存放两个线性表(a1, a2, a3…, am)和(b, b2, b3, …, bn)。试编 -//写一个函数, 将数组中两个顺序表的位置互换, 即将(b1,b2,b3,…, bn)放在(a1，a2, a3, …, am)的前面 - -void swaplist(List* A, int m, int n, int size) { - Reserve(A, 0, m + n - 1, size); - Reserve(A, 0, n - 1, size); - Reserve(A, n, m + n - 1, size); -} - -//P20.9线性表(a1,a2,a3,..,an)中的元素递增有序且按顺序存储于计算机内。 -//要求设计一算法, 完成用最少时间在表中查找数值为x的元素, 若找到则将其与后继元素位置相交换, -//若找不到则将其插入表中并使表中元素仍递增有序 - -//我的方法 大力出奇迹 -void CheckAndSwap(List* A, int x) { - bool flag = true; - _for(i, 0, A->length) { - if (A->data[i] == x) { - swap(&A->data[i], &A->data[i + 1]); - printf("Find it\n"); - flag = false; - break; - } - } - if (flag) { - printf("Not find\n"); - int i = 0; - for (i = 0; i < A->length && A->data[i] <= x; i++); - if (i < A->length) - { - for (int j = A->length; j >= i; j--) { - A->data[j + 1] = A->data[j]; - } - } - A->data[i] = x; - A->length += 1; - } -} - -//答案方法 折半查找 -void SearchExchangeInsert(List* A, int x) { - int low = 0, mid = A->length / 2, high = A->length - 1; - bool Isfind = false; - while (low <= high) { - mid = (low + high) / 2; - if (x == A->data[mid]) { - Isfind = true; - printf("Find\n"); - break; - } - if (x > A->data[mid]) { - low = mid + 1; - } - if (x < A->data[mid]) { - high = mid - 1; - } - } - if (Isfind && mid != A->length - 1) { - printf("Swap\n"); - swap(&A->data[mid], &A->data[mid + 1]); - } - if (!Isfind) { - int i = 0; - printf("%d", low); - for (i = A->length - 1; i > high; i--) { - A->data[i + 1] = A->data[i]; - } - A->data[i + 1] = x; - A->length += 1; - } -} - -//P20.10 -//〖2010统考真题〗设将n(n > 1)个整数存放到一维数组R中。设计一个在时间和空间两 -//方面都尽可能高效的算法。将R中保存的序列循环左移p(0length - 1, MAX); - Reserve(list, 0, list->length - 1, MAX); -} - -//P20.11 -//〖2011统考真题】一个长度为L(L≥1)的升序序列S, 处在第[L/2]个位置的数称为S的 -//中位数。例如, 若序列S1 = (1, 13, 15, 17, 19), 则S1的中位数是15, 两个序列的中位数 -//是含它们所有元素的升序序列的中位数。例如, 若S2 = (2, 4, 6, 8, 20), 则S1和S2的中位 -//数是11。现在有两个等长升序序列A和B, 试设计一个在时间和空间两方面都尽可能 -//高效的算法, 找出两个序列A和B的中位数。要求 -//1)给出算法的基本设计思想 -//2)根据设计思想, 采用C或C++或Java语言描述算法, 关键之处给出注释 -//3)说明你所设计算法的时间复杂度和空间复杂度 - -//改编的归并排序 -int FindMid(List A, List B) { - int i = 0, j = 0; - while (i + j != (A.length + B.length) / 2 - 1) { - if (A.data[i] <= B.data[j]) { - i++; - } - else { - j++; - } - } - return A.data[i] < B.data[j] ? A.data[i] : B.data[j]; -} - -//P20.12 -//〖2013统考真题〗已知一个整数序列A = (a0, a1…, an-1), 其中0 ≤ ai < n(0 ≤i < n)。若 -// 存在 ap1 = ap2 = ... = apm = x 且 m > n / 2 (0≤ Pk < n,1≤ k ≤ m), 则称x为A的主元素。 -// 例如A=(0,5,5,3,5,7,5,5)，则5为主元素; 又如A = (0, 5, 5, 3, 5, 1, 5, 7), 则A中没有主 -// 元素。假设A中的n个元素保存在一个一维数组中, 请设计一个尽可能高效的算法, 找 -// 出A的主元素。若存在主元素, 则输岀该元素; 否则输出 - 1。要求: -//1)给出算法的基本设计思想 -//2)根据设计思想, 采用C或C十或Java语言描述算法, 关键之处给出注释 -//3)说明你所设计算法的时间复杂度和空间复杂度 - -//次优算法，用空间换时间，类似桶排 -int FindMain(List A, int n) { - int m_num = 0; - int* Box = new int[n](); //动态申请一个大小为n的数组 - _for(i, 0, A.length) { - Box[A.data[i]]++; - } - _for(i, 0, n) { - if (Box[i] > n / 2) { - return A.data[i]; - } - } - return -1; -} - -//P21.13 -//〖2018统考真题】给定一个含n(n≥1)个整数的数组, 请设计一个在时间上尽可能高效 -//的算法, 找出数组中未出现的最小正整数。例如, 数组{ -5,3,2,3 }中未出现的最小正整数 -//是1; 数组{ 1,2,3 }中未出现的最小正整数是4。要求 -//1)给出算法的基本设计思想 -//2)根据设计思想, 采用C或C艹语言描述算法, 关键之处给出注释 -//3)说明你所设计算法的时间复杂度和空间复杂度 - -//用空间换时间 -int FindMinNum(List A, int n) { - int* Box = new int[n + 1](); - _for(i, 0, A.length) { - if (A.data[i] > 0 && A.data[i] <= n) - Box[A.data[i]]++; - } - _for(i, 1, n) { - if (!Box[i]) - return i; - } - return -1; -} - -int main() -{ - List list1; - List list2; - list1.length = 4; - list2.length = 5; - int Data1[] = { 1,3,2,3 }; - int Data2[] = { 22,43,64,87,180 }; - - for (int i = 0; i < list1.length; i++) - list1.data[i] = Data1[i]; - - for (int i = 0; i < list2.length; i++) - list2.data[i] = Data2[i]; - - printf("%d", FindMinNum(list1, 4)); -} \ No newline at end of file