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krahets
2023-10-27 23:48:10 +08:00
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commit 368cbf4261
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docs/chapter_backtracking/subset_sum_problem.md
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@@ -389,38 +389,34 @@ comments: true
```c title="subset_sum_i_naive.c"
/* 回溯算法：子集和 I */
void backtrack(vector *state, int target, int total, vector *choices, vector *res) {
void backtrack(int target, int total, int *choices, int choicesSize) {
// 子集和等于 target 时，记录解
if (total == target) {
vector *tmpVector = newVector();
for (int i = 0; i < state->size; i++) {
vectorPushback(tmpVector, state->data[i], sizeof(int));
for (int i = 0; i < stateSize; i++) {
res[resSize][i] = state[i];
}
vectorPushback(res, tmpVector, sizeof(vector));
resColSizes[resSize++] = stateSize;
return;
}
// 遍历所有选择
for (size_t i = 0; i < choices->size; i++) {
for (int i = 0; i < choicesSize; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + *(int *)(choices->data[i]) > target) {
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
vectorPushback(state, choices->data[i], sizeof(int));
state[stateSize++] = choices[i];
// 进行下一轮选择
backtrack(state, target, total + *(int *)(choices->data[i]), choices, res);
backtrack(target, total + choices[i], choices, choicesSize);
// 回退：撤销选择，恢复到之前的状态
vectorPopback(state);
stateSize--;
}
}
/* 求解子集和 I（包含重复子集） */
vector *subsetSumINaive(vector *nums, int target) {
vector *state = newVector(); // 状态（子集）
int total = 0; // 子集和
vector *res = newVector(); // 结果列表（子集列表）
backtrack(state, target, total, nums, res);
return res;
void subsetSumINaive(int *nums, int numsSize, int target) {
resSize = 0; // 初始化解的数量为0
backtrack(target, 0, nums, numsSize);
}
```
@@ -867,40 +863,38 @@ comments: true
```c title="subset_sum_i.c"
/* 回溯算法：子集和 I */
void backtrack(vector *state, int target, vector *choices, int start, vector *res) {
void backtrack(int target, int *choices, int choicesSize, int start) {
// 子集和等于 target 时，记录解
if (target == 0) {
vector *tmpVector = newVector();
for (int i = 0; i < state->size; i++) {
vectorPushback(tmpVector, state->data[i], sizeof(int));
for (int i = 0; i < stateSize; ++i) {
res[resSize][i] = state[i];
}
vectorPushback(res, tmpVector, sizeof(vector));
resColSizes[resSize++] = stateSize;
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < choices->size; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (target - *(int *)(choices->data[i]) < 0) {
for (int i = start; i < choicesSize; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
vectorPushback(state, choices->data[i], sizeof(int));
state[stateSize] = choices[i];
stateSize++;
// 进行下一轮选择
backtrack(state, target - *(int *)(choices->data[i]), choices, i, res);
backtrack(target - choices[i], choices, choicesSize, i);
// 回退：撤销选择，恢复到之前的状态
vectorPopback(state);
stateSize--;
}
}
/* 求解子集和 I */
vector *subsetSumI(vector *nums, int target) {
vector *state = newVector(); // 状态（子集）
qsort(nums->data, nums->size, sizeof(int *), comp); // 对 nums 进行排序
int start = 0; // 子集和
vector *res = newVector(); // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
void subsetSumI(int *nums, int numsSize, int target) {
qsort(nums, numsSize, sizeof(int), cmp); // 对 nums 进行排序
int start = 0; // 遍历起始点
backtrack(target, nums, numsSize, start);
}
```
@@ -1383,46 +1377,43 @@ comments: true
```c title="subset_sum_ii.c"
/* 回溯算法：子集和 II */
void backtrack(vector *state, int target, vector *choices, int start, vector *res) {
void backtrack(int target, int *choices, int choicesSize, int start) {
// 子集和等于 target 时，记录解
if (target == 0) {
vector *tmpVector = newVector();
for (int i = 0; i < state->size; i++) {
vectorPushback(tmpVector, state->data[i], sizeof(int));
for (int i = 0; i < stateSize; i++) {
res[resSize][i] = state[i];
}
vectorPushback(res, tmpVector, sizeof(vector));
resColSizes[resSize++] = stateSize;
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < choices->size; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - *(int *)(choices->data[i]) < 0) {
for (int i = start; i < choicesSize; i++) {
// 剪枝一：若子集和超过 target ，则直接跳过
if (target - choices[i] < 0) {
continue;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && *(int *)(choices->data[i]) == *(int *)(choices->data[i - 1])) {
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
vectorPushback(state, choices->data[i], sizeof(int));
state[stateSize] = choices[i];
stateSize++;
// 进行下一轮选择
backtrack(state, target - *(int *)(choices->data[i]), choices, i + 1, res);
backtrack(target - choices[i], choices, choicesSize, i + 1);
// 回退：撤销选择，恢复到之前的状态
vectorPopback(state);
stateSize--;
}
}
/* 求解子集和 II */
vector *subsetSumII(vector *nums, int target) {
vector *state = newVector(); // 状态（子集）
qsort(nums->data, nums->size, sizeof(int *), comp); // 对 nums 进行排序
int start = 0; // 子集和
vector *res = newVector(); // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
void subsetSumII(int *nums, int numsSize, int target) {
// 对 nums 进行排序
qsort(nums, numsSize, sizeof(int), cmp);
// 开始回溯
backtrack(target, nums, numsSize, 0);
}
```