[Leetcode] Combination Sum

由于每个元素我可以出现的次数没有限制，我们可以在使用某个元素的时候进行计算一下最多的个数，进行枚举。

同样的道理，在每一层递归只考虑一个元素，并且在当前sum大于目标sum的时候进行剪枝。

 

import java.util.*;

public class Solution {
    public void dp(int[] candidates, int index, int target,List<List<Integer>> lists,
    List<Integer> tmplist, int tmpsum){
        if(tmpsum==target) {
            List<Integer> l = new LinkedList<Integer>();
            l.addAll(tmplist);
            lists.add(l);
            return;
        }
        if(tmpsum>target||index==candidates.length) return;
        int maxnum=(target-tmpsum)/candidates[index];
        for(int i=0;i<maxnum;i++){
            tmplist.add(candidates[index]);
            dp(candidates,index+1,target,lists,tmplist,tmpsum+candidates[index]*(i+1));
        }
        for(int i=0;i<maxnum;i++){
            tmplist.remove(tmplist.size()-1);
        }
        dp(candidates,index+1,target,lists,tmplist,tmpsum);
    }
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        // if the candidates are not sorted ,then we need to sort each conditioned 
        // tmplist. Because it need the tmplist should be sorted.
        Arrays.sort(candidates);
        List<List<Integer>> lists =new LinkedList<List<Integer>> ();
        List<Integer> tmplist = new LinkedList<Integer>();
        dp(candidates,0,target,lists,tmplist,0);
        return lists;
    }
}