LeetCode刷题笔记-39.组合总和

C语言

/**
 * 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 dfs(int *nums, int numsSize, int *cs, int sum, int ***ret, int *returnSize, int **colSize, int curi) {
  int i;
  
  if (sum < 0) {
    return;
  }
  
  if (sum == 0) {
    int tcs = 0;
    int tail = 0;
    int j;
    int ci = 0;
    //TODO 遍历所有数量
    for (i = 0; i < numsSize; i++) {
      tcs += cs[i];
    }
    //开辟list空间 cols空间
    (*returnSize)++;
    tail = (*returnSize) - 1; 
    *ret = realloc(*ret, (*returnSize) * (sizeof(int *)));
    *colSize = realloc(*colSize, (*returnSize) * sizeof(int));
    (*ret)[tail] = calloc(tcs, sizeof(int));
    (*colSize)[tail] = tcs;
    //填值
    for (i = 0; i < numsSize; i++) {
      for (j = 0; j < cs[i]; j++) {
        (*ret)[tail][ci] = nums[i];
        ci++;
      }
    }
    return;
  }
  
  for (i = 0; i < numsSize; i++) {
    if (i < curi)
      continue;
    cs[i]++;
    dfs(nums, numsSize, cs, sum - nums[i], ret, returnSize, colSize, i);
    cs[i]--;
  }
  return;
}

int** combinationSum(int* nums, int numsSize, int target, int* returnSize, int** returnColumnSizes) {
  int **ret = NULL;
  int *cs = NULL;
  *returnSize = 0;
  ret = calloc(0, sizeof(int *));
  *returnColumnSizes = calloc(0, sizeof(int));
  
  cs = calloc(numsSize, sizeof(int));
  
  dfs(nums, numsSize, cs, target, &ret, returnSize, returnColumnSizes, 0);
  
  return ret;
}

注意点

1. 回溯结束后记得把增加了的计数减1；
2. 注意去掉重复项，如果每次都是全遍历，举个例子，如同用例1:会出现【2，2，3】【2，3，2】【3，2，2】会满足递归条件，所以本人在题解中加入了curi来控制重复情况；

结果

题目