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给你一个整数数组 nums ，数组中的元素 互不相同 。返回该数组所有可能的子集（幂集）。

解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。

思路一：回溯

void backtracking(int* nums, int numsSize, int** res, int* returnSize, int** returnColumnSizes, int* path, int pathSize, int startIndex) {res[*returnSize] = (int*)malloc(sizeof(int) * pathSize);memcpy(res[*returnSize], path, sizeof(int) * pathSize);(*returnColumnSizes)[*returnSize] = pathSize;(*returnSize)++;for (int i = startIndex; i < numsSize; i++) {path[pathSize] = nums[i];backtracking(nums, numsSize, res, returnSize, returnColumnSizes, path, pathSize + 1, i + 1);}
}int** subsets(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {*returnSize = 0;*returnColumnSizes = (int*)malloc(sizeof(int) * 10001);int** res = (int**)malloc(sizeof(int*) * 10001);int* path = (int*)malloc(sizeof(int) * numsSize);backtracking(nums, numsSize, res, returnSize, returnColumnSizes, path, 0, 0);return res;
}

分析：

本题与上一题相似，利用回溯算法将数组内子集全部列出即可，path[pathSize] = nums[i];

backtracking(nums, numsSize, res, returnSize, returnColumnSizes, path, pathSize + 1, i + 1);将子集全部列出，最后返回res

总结：

本题考察回溯的应用，将子集按顺序全部列出即可解决