[leetcode-39]Combination Sum(java)

问题描述：

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

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.

Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7,

A solution set is:

[7]

[2, 2, 3]

分析：这道题如果不用DFS，如果不用递归+剪枝，那么需要很多很多细节来考虑，我之前觉得递归太耗资源，所以希望直接做，但是还是理不清思路。而且递归的方法，就好像已经是一个定式了，只要按照一定模式去做，就可以肯定返回正确答案了。

代码如下：344ms

[code]public class Solution {
    private void solve(List<List<Integer>> res,int currentIndex,int count,List<Integer> tmp,int[] candidates,int target){
        if(count>=target) {
            if(count==target)
                res.add(new LinkedList<>(tmp));
            return;
        }

        for(int i = currentIndex;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);
        }
    }
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new LinkedList<List<Integer>>();
        List<Integer> tmp = new LinkedList<>();
        Arrays.sort(candidates);

        solve(res,0,0,tmp,candidates,target);
        return res;
    }
}