LeetCode 40题：组合总和2

题目

给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用 一次 。

注意：解集不能包含重复的组合。 

示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

示例 2:

输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]

提示:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

---

 

代码

#include
#include
#include

int comp(const void *a, const void *b);
int combine(int* candidates, int candidatesSize, int target,int**res,int ind);
int** combinationSum2(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes);

int size=0;
int len=0;
int *com;
int *columnsize;

int main()
{
    int candidates[5] = {2,5,2,1,2}, target = 5;
    int*t=(int*)malloc(sizeof(int)),**te=(int**)malloc(sizeof(int*));
    int**res=combinationSum2(candidates,5,target,t,te);
    printf("\n");
    for(int i=0;i<*t;i++)
    {
        for(int j=0;jtarget)
    {
        *returnSize=0;
        returnColumnSizes[0][0]=0;
        return NULL;
    }
    int t=combine(candidates,candidatesSize, target,res,0);
    *returnSize=size;
    returnColumnSizes[0]=columnsize;
    return res;
}

int combine(int* candidates, int candidatesSize, int target,int**res,int ind)
{
    if(target<0)
    {
        return 1;
    }
    else if(target==0)
    {
        size++;
        columnsize[size-1]=len;
        int*temp=(int*)malloc(sizeof(int)*len);
        for(int i=0;i