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:

复制代码

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

cs 复制代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

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

int search(int target,int**res,int*combine,int ind,int*candidates,int candidatesSize);
int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes);

int main()
{
    int candidates[4] = {2,3,5}, target = 8;
    int *returnSizes=(int*)malloc(sizeof(int)),**returnColumnSize=(int**)malloc(sizeof(int*));
    returnColumnSize[0]=(int*)malloc(sizeof(int));
    int **res=combinationSum(candidates,3,target,returnSizes,returnColumnSize);
    for(int i=0;i<*returnSizes;i++)
    {
        for(int j=0;j<returnColumnSize[0][i];j++)
        {
            printf("%d ",res[i][j]);
        }
        printf("\n");
    }
}

int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes)
{
    columnsize=(int*)malloc(sizeof(int));
    int**res=(int**)malloc(sizeof(int*)*150);
    if(candidates[0]>target)
    {
        *returnSize=0;
        returnColumnSizes[0][*returnSize]=0;
        return NULL;
    }
    int*combine=(int*)malloc(sizeof(int)*(target/candidates[0]+1)); 
    int t= search(target,res,combine,0,candidates,candidatesSize);
    *returnSize=size;
    returnColumnSizes[0]=columnsize;
    return res;
}

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

官方提供

尝试用「搜索回溯」的方法来解决。当然，搜索回溯的过程一定存在一些优秀的剪枝方法来使得程序运行得更快，而这里只给出了最朴素不含剪枝的写法

cs 复制代码

int candidatesSize_tmp;

int ansSize;

int combineSize;

int* ansColumnSize;

void dfs(int* candidates, int target, int** ans, int* combine, int idx) {
    if (idx == candidatesSize_tmp) {
        return;
    }
    if (target == 0) {
        int* tmp = malloc(sizeof(int) * combineSize);
        for (int i = 0; i < combineSize; ++i) {
            tmp[i] = combine[i];
        }
        ans[ansSize] = tmp;
        ansColumnSize[ansSize++] = combineSize;
        return;
    }
    // 直接跳过
    dfs(candidates, target, ans, combine, idx + 1);
    // 选择当前数
    if (target - candidates[idx] >= 0) {
        combine[combineSize++] = candidates[idx];
        dfs(candidates, target - candidates[idx], ans, combine, idx);
        combineSize--;
    }
}

int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes) {
    candidatesSize_tmp = candidatesSize;
    ansSize = combineSize = 0;
    int** ans = malloc(sizeof(int*) * 1001);
    ansColumnSize = malloc(sizeof(int) * 1001);
    int combine[2001];
    dfs(candidates, target, ans, combine, 0);
    *returnSize = ansSize;
    *returnColumnSizes = ansColumnSize;
    return ans;
}

作者：力扣官方题解

链接：https://leetcode.cn/problems/combination-sum/solutions/406516/zu-he-zong-he-by-leetcode-solution/

来源：力扣（LeetCode）

LeetCode 39题：组合总和

题目

代码

官方提供