040 - Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, … ,a_k) must be in non-descending order. (ie, a₁ ≤a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]

[1, 1, 6]

同39题，只是这次数列中的值不能重复使用，用一个数组来标记是否已使用过

int mycompar(const void *a, const void *b){int *x = (int *)a;int *y = (int *)b;if (*x > *y) return 1;return *x == *y? 0 : -1;}void dofun(int *nlist, int nsize, int *nmark, int left, int target, int **ret, int *col, int *retsize, int *tmp, int *tmpsize){int i;if (target==0) {//tmp[(*tmpsize)++] = nlist[0];int *line = (int *)calloc(sizeof(int), *tmpsize);col[*retsize] = *tmpsize;memcpy(line, tmp, *tmpsize * sizeof(int));ret[*retsize] = line;(*retsize)++;return;}if (target < nlist[0]) return;for (i = left + 1; i < nsize; i++) {if (nmark[i]) continue;nmark[i] = 1;tmp[(*tmpsize)++] = nlist[i];int left = i;dofun(nlist, nsize, nmark, left, target - nlist[i], ret, col, retsize, tmp, tmpsize);(*tmpsize)--;nmark[i] = 0;while (nlist[i+1] == nlist[i]) i++;}}int **combinationSum2(int *nlist, int nsize, int target, int **col, int *retsize){qsort(nlist, nsize, sizeof(int), mycompar);int **ret = (int **)calloc(sizeof(int *), 8192);*col = (int *)calloc(sizeof(int), 8192);int tmp[8192] = {0};int tmpsize = 0;int nmark[8192] = {0};int left = -1;dofun(nlist, nsize, nmark, left, target, ret, *col, retsize, tmp, &tmpsize);return ret;}