【LeetCode】#39组合总和(Combination Sum)

加粗样式

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。
说明：
所有数字（包括 target）都是正整数。
解集不能包含重复的组合。

示例

示例 1:

输入: candidates = [2,3,6,7], target = 7,
所求解集为:
[
[7],
[2,2,3]
]

示例 2:

输入: candidates = [2,3,5], target = 8,
所求解集为:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

Description

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

解法

class Solution{
	public List<List<Integer>> combinationSum(int[] candidates, int target){
		List<List<Integer>> res = new LinkedList<>();
		List<Integer> tmp = new LinkedList<>();
		Arrays.sort(candidates);
		solve(res, 0, 0, tmp, candidates, target);
		return res;
	}
	private void solve(List<List<Integer>> res, int currentIndex, 
		int count, List<Integer> tmp, int[] candidates, int target){
		if(count>=target){
			if(count==targer)
				res.add(new LinkedList<>(tmp);
			return;
		}
		for(int i=candidates; i<candidates.length; i++){
			if(count+candidates[i]>target)
				break;
			tmp.add(candidates[i]);
			solve(res, i, count+candidates[i], tmp, candidates, target);
			tmp.remove(tmp.size()-1);
		}
	}
}