39.组合总和.md

39. 组合总和

url

题目

给定一个无重复元素的数组 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 中的数字可以无限制重复被选取。

方法

code

js

let combinationSum = (candidates, target) => {
    let dfs = (nums, target, start, list) => {
        if (target === 0) {
            ret.push(list.slice());
            return;
        }
        for (let i = start; i < nums.length; i++) {
            if (nums[i] <= target) {
                list.push(nums[i]);
                dfs(nums, target - nums[i], i, list);
                list.pop();
            }
        }
    }
    let ret = []
    if (candidates === null || candidates.length === 0)
        return ret;
    dfs(candidates, target, 0, []);
    return ret;
}

go

func combinationSum(candidates []int, target int) [][]int {
	var res [][]int
	if len(candidates) == 0 {
		return res
	}
	var dfs func(nums []int, target int, start int, list []int)
	dfs = func(nums []int, target int, start int, list []int) {
		if target == 0 {
			res = append(res, append([]int{}, list...))
			return
		}
		for i := start; i < len(nums); i++ {
			if nums[i] <= target {
				list = append(list, nums[i])
				dfs(nums, target - nums[i], i, list)
				list = list[0:len(list) - 1]
			}
		}
	}
	dfs(candidates, target, 0, []int{})
	return res
}

java

class Solution {
    List<List<Integer>> ret = new ArrayList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if (candidates == null || candidates.length == 0)
            return ret;
        dfs(candidates, target, 0, new ArrayList<Integer>());
        return ret;
    }
    public void dfs(int[] nums, int target, int start, ArrayList<Integer> list){
        if (target == 0) {
            ret.add(new ArrayList(list));
            return;
        }
        for (int i = start; i < nums.length; i++){
            if (nums[i] <= target) {
                list.add(nums[i]);
                dfs(nums, target - nums[i], i, list);
                list.remove(list.size() - 1);
            }
        }
    }
}