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<ArrayList<Integer>> combinationSum2(int[] num, int target) { 3 ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>(); 4 ArrayList<Integer> tmp=new ArrayList<Integer>(); 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<ArrayList<Integer>> res,ArrayList<Integer> tmp){ 10 if(target<0)return; 11 if(target==0 && !res.contains(tmp)){ 12 res.add(new ArrayList<Integer>(tmp)); 13 return ; 14 } 15 for(int i=start;i<num.length;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 }