leetcode------Combination Sum II

标题: Combination Sum II
通过率: 25.1%
难度: 中等

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

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

 

与第一个版本不同的是set中的元素不能重复使用,那么递归时起始位置就要+1操作了代码如下:

 1 public class Solution {
 2     public ArrayList> combinationSum2(int[] num, int target) {
 3         ArrayList> res=new ArrayList>();
 4         ArrayList tmp=new ArrayList();
 5         Arrays.sort(num);
 6         dfs(num,target,0,res,tmp);
 7         return res;
 8     }
 9     public void dfs(int[] num, int target,int start,ArrayList> res,ArrayList tmp){
10         if(target<0)return;
11         if(target==0 && !res.contains(tmp)){
12             res.add(new ArrayList(tmp));
13             return ;
14         }
15         for(int i=start;i){
16             tmp.add(num[i]);
17             dfs(num,target-num[i],i+1,res,tmp);
18             tmp.remove(tmp.size()-1);
19         }
20     }
21 }

 

转载于:https://www.cnblogs.com/pkuYang/p/4354384.html

你可能感兴趣的:(leetcode------Combination Sum II)