40-m-Combination Sum II

求指定和的组合，但个个数只能用一次。求和一的变种，有了一的经验，二式可以沿用一式的backtracking格式，由于每个数只能用一次，那么控制参数i每次递增1就可以了。另一个要注意的是可能有重复项而重复项可能会构成解，例如【1，1，2，5，6】target = 8，【1，1，6】是解但会出现2组【1，2，5】，因此要过滤掉用过一次的重复项，我的方法是在本层递归后将重复项跃过（因为是在递归后跃进，所以能保证是用过一次的）。

ac代码如下：

void SortThreeSum(int num[], int n, int l, int h){    if (l >= h)        return;    int low = l, high = h;    int temp = num[low];    while (low < high) {        while (temp < num[high] && low < high)            high--;        num[low] = num[high];        while (temp >= num[low] && low < high)            low++;        num[high] = num[low];    }    num[low] = temp;    SortThreeSum(num, n, l, low - 1);    SortThreeSum(num, n, low + 1, h);}void cbs2(int *candidates, int candidatesSize, int target, int **columnSizes, int *returnSize, int indexi, int ii, int **path, int sum, int ***result) {    if (sum == target) {        columnSizes[0][*returnSize] = indexi;        printf("indexi-->%d\n", indexi);        *returnSize += 1;        *columnSizes = (int *)realloc(*columnSizes, sizeof(int) * (*returnSize + 1));                for (int i = 0; i < columnSizes[0][*returnSize - 1]; i++) {            (*result)[*returnSize - 1][i] = path[0][i];            // printf("%d,", path[0][i]);        }//        printf("\n");        int n = target / candidates[0] + 1;        (*result) = (int **)realloc(*result, sizeof(int *) * (*returnSize + 1));        (*result)[*returnSize] = (int *)malloc(sizeof(int) * n);//        memset(*path, 0, n);        return;    }    for (int i = ii; i < candidatesSize; i++) {//        sum += candidates[i]; //注意sum不能写在循环里，而应该写在形参里，以此保持本层递归中sum的对应        if (sum + candidates[i] <= target) {            path[0][indexi] = candidates[i];            cbs2(candidates, candidatesSize, target, columnSizes, returnSize, indexi + 1, i + 1, path, sum + candidates[i], result);        }        else            return;        while (candidates[i] == candidates[i + 1] && i < candidatesSize - 1)            i++;    }}int** combinationSum2(int* candidates, int candidatesSize, int target, int** columnSizes, int* returnSize) {    int **result = NULL;    SortThreeSum(candidates, candidatesSize, 0, candidatesSize - 1);    int n = target / candidates[0] + 1;    result = (int **)malloc(sizeof(int *));    *result = (int *)malloc(sizeof(int) * n);    int ***threerp = &result;        *columnSizes = (int *)malloc(sizeof(int));    int ***threecp = &columnSizes;        int **path = (int **)malloc(sizeof(int *));    *path = (int *)malloc(sizeof(int) * n);    memset(*path, 0, sizeof(int) * n);        int indexi = 0, sum = 0;    int ii = 0;        cbs2(candidates, candidatesSize, target, columnSizes, returnSize, indexi, ii, path, sum, threerp);        return result;}