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高端网站建设注意,玉林市建设局网站,做了网站应该如何推广,北京城建十建设工程有限公司网站前言思路及算法思维，指路代码随想录。题目来自 LeetCode。 day26， 休息的周末~ day 27，周一，库存没了，哭死~ 题目详情 [39] 组合总和题目描述 39 组合总和解题思路前提：组合的子集问题&…

前言

思路及算法思维，指路代码随想录。
题目来自 LeetCode。

day26，休息的周末~
day 27，周一，库存没了，哭死~

题目详情

[39] 组合总和

题目描述

39 组合总和

解题思路

前提：组合的子集问题，统一元素可以重复选取
思路：回溯 + 剪枝。
重点：剪枝的前提是数组已排序。

代码实现

C语言

回溯 + 未排序剪枝

/*** Return an array of arrays of size *returnSize.* The sizes of the arrays are returned as *returnColumnSizes array.* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().*/void backtracing(int* candidates, int candidatesSize, int target, int index, int *nums, int numsSize, int ***ans, int* returnSize, int** returnColumnSizes)
{// 退出条件if (0 == target){*ans = (int **)realloc(*ans, sizeof(int *) * ((*returnSize) + 1));(*ans)[*returnSize] = (int *)malloc(sizeof(int) * (numsSize));for (int i = 0; i < numsSize; i++){(*ans)[*returnSize][i] = nums[i];}*returnColumnSizes = (int *)realloc(*returnColumnSizes, sizeof(int) * ((*returnSize) + 1));(*returnColumnSizes)[*returnSize] = numsSize;(*returnSize)++;return ;}for (int j = index; j < candidatesSize; j++){if (target < candidates[j]){continue ;}// 递归nums[numsSize] = candidates[j];numsSize++;backtracing(candidates, candidatesSize, target - candidates[j], j, nums, numsSize, ans, returnSize, returnColumnSizes);// 回溯numsSize--;nums[numsSize] = 0;}return ;
}int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes) {// 判空if (candidatesSize == 0){return NULL;}// 输出int **ans = NULL;int nums[41];int index = 0;*returnSize = 0;printf("%d\n", target);backtracing(candidates, candidatesSize, target, 0, nums, 0, &ans, returnSize, returnColumnSizes);if (*returnSize == 0){return NULL;}return ans;
}

回溯 + 排序 + 剪枝

/*** Return an array of arrays of size *returnSize.* The sizes of the arrays are returned as *returnColumnSizes array.* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().*/int cmp(const void *p1, const void *p2)
{return *(int *)p1 > *(int *)p2;
}void backtracing(int* candidates, int candidatesSize, int target, int index, int *nums, int numsSize, int ***ans, int* returnSize, int** returnColumnSizes)
{// 退出条件if (0 == target){*ans = (int **)realloc(*ans, sizeof(int *) * ((*returnSize) + 1));(*ans)[*returnSize] = (int *)malloc(sizeof(int) * (numsSize));for (int i = 0; i < numsSize; i++){(*ans)[*returnSize][i] = nums[i];}*returnColumnSizes = (int *)realloc(*returnColumnSizes, sizeof(int) * ((*returnSize) + 1));(*returnColumnSizes)[*returnSize] = numsSize;(*returnSize)++;return ;}// 剪枝for (int j = index; (j < candidatesSize) && (target >= candidates[j]); j++){// 递归nums[numsSize] = candidates[j];numsSize++;backtracing(candidates, candidatesSize, target - candidates[j], j, nums, numsSize, ans, returnSize, returnColumnSizes);// 回溯numsSize--;nums[numsSize] = 0;}return ;
}int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes) {// 判空if (candidatesSize == 0){return NULL;}// 排序qsort(candidates, candidatesSize, sizeof(int), cmp);// 输出int **ans = NULL;int nums[41];int index = 0;*returnSize = 0;backtracing(candidates, candidatesSize, target, 0, nums, 0, &ans, returnSize, returnColumnSizes);if (*returnSize == 0){return NULL;}return ans;
}

[40] 组合总和II

题目描述

40 组合总和II

解题思路

前提：组合的子集问题，同一元素只能使用一次，但是结果不包含重复组合
思路：回溯 + 剪枝
重点：结果子集中排除重复组合，需要树形结构中，同一树层的相同的元素值不可重复选取，使用used数组实现去重。

代码实现

C语言

利用used数组 false，同一树层去重

/*** Return an array of arrays of size *returnSize.* The sizes of the arrays are returned as *returnColumnSizes array.* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().*/int cmp(const void *p1, const void *p2)
{return *(int *)p1 > *(int *)p2;
}void backtracing(int* candidates, int candidatesSize, int target, int index, int *nums, int numsSize, bool *used, int ***ans, int* returnSize, int** returnColumnSizes)
{// 退出条件if (0 == target){*ans = (int **)realloc(*ans, sizeof(int *) * ((*returnSize) + 1));(*ans)[*returnSize] = (int *)malloc(sizeof(int) * (numsSize));for (int i = 0; i < numsSize; i++){(*ans)[*returnSize][i] = nums[i];}*returnColumnSizes = (int *)realloc(*returnColumnSizes, sizeof(int) * ((*returnSize) + 1));(*returnColumnSizes)[*returnSize] = numsSize;(*returnSize)++;return ;}for (int j = index; (j < candidatesSize) && (target >= candidates[j]); j++){// 去重if ((j > 0) && (candidates[j] == candidates[j - 1]) && (used[j - 1] == false)){continue;}// 递归nums[numsSize] = candidates[j];used[j] = true;numsSize++;backtracing(candidates, candidatesSize, target - candidates[j], j + 1, nums, numsSize, used, ans, returnSize, returnColumnSizes);// 回溯numsSize--;used[j] = false;nums[numsSize] = 0;}return ;
}int** combinationSum2(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes) {// 判空if (candidatesSize == 0){return NULL;}// 排序qsort(candidates, candidatesSize, sizeof(int), cmp);// 输出int **ans = NULL;int nums[100] = {0};bool used[100] = {false};int index = 0;*returnSize = 0;backtracing(candidates, candidatesSize, target, 0, nums, 0, used, &ans, returnSize, returnColumnSizes);if (*returnSize == 0){return NULL;}return ans;
}