简历投递一天

2026-02-03 18:44:19 +08:00 · 2021-08-23 22:53:15 +08:00
parent 4ed6ee6517
commit fb01b19088
7 changed files with 304 additions and 13 deletions
--- a/code_segment/binary_find.cpp
+++ b/code_segment/binary_find.cpp
@@ -0,0 +1,99 @@
+class Solution {
+public:
+    double findone(vector<int>&nums){
+        int n = nums.size();
+        if(n==1)return nums[0];
+        if(n%2==1){
+            
+            return nums[(n-1)/2];
+        }
+        else{
+            // cout<< nums[(n-1)/2]<<endl;
+            return (nums[(n-1)/2]+nums[(n-1)/2+1])/2;
+        }
+    }
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        int n1 = nums1.size(),n2=nums2.size();
+        int i=0,j=0,n=n1+n2;
+        while(i+j<n/2){
+            if(nums1[i]<nums2[j] && i<n1 || j>=n2){
+                i++;
+            }
+            else if(j<n2){
+                j++;
+            }
+        }
+
+    }
+    double findMedianSortedArrays2(vector<int>& nums1, vector<int>& nums2) {
+        // 一种新的二分查找。每次排除掉两个数组的1/4。直接扫描两遍也行？这个数据规模，爆搜吧。
+        int n1 = nums1.size(),n2=nums2.size();
+        int l1=0,m1=0,r1=n1-1;
+        int l2=0,m2=0,r2=n2-1;
+        m1 = (l1+r1)/2;
+        m2 = (l2+r2)/2;
+        int n = n1+n2;
+        int *m,*l,*r;
+        // 特殊情况的讨论
+
+        if(n1==0){
+            return findone(nums2);
+        }
+        if(n2==0){
+            return findone(nums1);
+        }
+        if(n1==1){
+            nums2.push_back(nums1[0]);
+            sort(nums2.begin(),nums2.end());
+            return findone(nums2);
+        }
+        if(n2==1){
+            nums1.push_back(nums2[0]);
+            sort(nums1.begin(),nums1.end());
+            // for(auto a : nums1){
+            //     cout<<a<<" ";
+            // }
+            return findone(nums1);
+        }
+        // 保证左边的元素小于等于右
+        while(true){
+            // 终止搜索的条件。l1和l2不需要相邻。都完成二分搜索后，只剩下两个数。
+            if(r1-l1<=1 && r2-l2<=1){
+                cout<<l1<<" "<<r1<<" "<<l2<<" "<<r2<<endl;
+                vector<int> temp{nums1[l1],nums1[r1],nums2[l2],nums2[r2]};
+                sort(temp.begin(),temp.end());
+                int left = l1+l2;
+                int right = n-r1-r2-2;
+                int index = ((right-left)+3)/2;
+
+                if(n%2==1){
+                    return temp[index];
+                }
+                else{
+                    return (temp[index]+temp[index+1])/2.0;
+                }
+            }
+
+            // 继续搜索，排除掉1/4的方法.暂不考虑相等
+            // 选择做二分的数组
+            if(nums1[m1]<nums2[m2] && m1+m2<n/2 && r1-l1>1 || nums1[m1]>nums2[m2] && m1+m2>n/2 && r1-l1>1){
+                m = &m1;
+                l=&l1;
+                r=&r1;
+            }else if(r2-l2>1){
+                m = &m2;
+                l = &l2;
+                r = &r2;
+            }
+            if(m1+m2<n/2){
+                *l = *m;
+                *m = (*l+*r)/2;
+            }
+            else{
+                *r = *m;
+                *m = (*l+*r)/2;
+            }
+        }
+        return 1;
+    }
+};