leetcode 78 子集 C语言实现

题目

leetcode 78
给定一组不含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。
说明：解集不能包含重复的子集。
示例:
输入: nums = [1,2,3]
输出:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

代码

回溯法

//nums[c],arr_t[ind],n是代表当前要形成的数组长度(1-numsSize)
int dfs(int *nums, int c, int numsSize, int arr_t[], int ind, int n, int **arr, int *returnSize, int **returnColumnSizes) {
    if (ind == n) {
        arr[*returnSize] = (int *)malloc(sizeof(int) * n);
        for (int i = 0; i < n; i++) {
            arr[*returnSize][i] = arr_t[i];
        }
        (*returnColumnSizes)[*returnSize] = n;
        (*returnSize)++;
        for (int i = 1; i <= n; i++) {
            if (nums[numsSize - i] != arr_t[n - i]) return 0;
        }
        if (n == numsSize) return 0;
        return 1;
    }
    for (int i = c; i < numsSize; i++) {
        arr_t[ind] = nums[i];
        int flag = dfs(nums, i + 1, numsSize, arr_t, ind + 1, n, arr, returnSize, returnColumnSizes);
        if (flag == 1) {
            n++;
            i = -1;
            ind = 0;
        }
    }
    return 0;
}
int** subsets(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
    long int len = 1;
    int count = 0;
    while (count < numsSize) {
        len *= 2;
        count++;
    }
    int **arr = (int **)malloc(sizeof(int *) * len);
    (*returnColumnSizes) = (int *)malloc(sizeof(int) * len);
    int arr_t[numsSize];
    *returnSize = 0;
    arr[*returnSize] = (int *)malloc(sizeof(int) * 1);
    (*returnColumnSizes)[*returnSize] = 0;
    (*returnSize)++;
    dfs(nums, 0, numsSize, arr_t, 0, 1, arr, returnSize, returnColumnSizes);
    return arr;
}