leetcode do question notes 78 subset

You are given an array of integers  nums , the elements in the array  are different from each other  . Returns all possible subsets (power sets) of this array.

A solution set  cannot  contain duplicate subsets. You can   return the solution sets in any order .

Idea 1: Backtracking

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;
}

analyze:

This question is similar to the previous question, just use the backtracking algorithm to list all the subsets in the array, path[pathSize] = nums[i];

backtracking(nums, numsSize, res, returnSize, returnColumnSizes, path, pathSize + 1, i + 1); list all the subsets, and finally return res

Summarize:

This question examines the application of backtracking, and it can be solved by listing all the subsets in order