【问题】
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]
【分析】
1.与题一思路类似,只是在递归处参数改动
【算法实现】
//v1 public class Solution { List<List<Integer>> res; List<Integer> temp; public List<List<Integer>> combinationSum2(int[] num, int target) { res=new ArrayList<List<Integer>>(); temp=new ArrayList<Integer>(); Arrays.sort(num); getCombination(num,target,0,0); HashSet h = new HashSet(res); //去除重复的list res.clear(); res.addAll(h); return res; } public void getCombination(int[] num,int target,int sum,int level) { if(sum==target) { res.add(new ArrayList<Integer>(temp)); return; } if(sum>target||level==num.length) return; for(int i=level;i<num.length;i++) { sum+=num[i]; temp.add(num[i]); getCombination(num,target,sum,i+1); temp.remove(temp.size()-1); sum-=num[i]; } } }
//v2
public class Solution { List<List<Integer>> result; List<Integer> solu; public List<List<Integer>> combinationSum2(int[] num, int target) { result = new ArrayList<>(); solu = new ArrayList<>(); Arrays.sort(num); getComboSum(num, target, 0, 0); return result; } public void getComboSum(int []num, int target, int sum, int level){ if(sum==target){ result.add(new ArrayList<>(solu)); return; } if(sum>target) return; for(int i=level;i<num.length;i++){ sum+=num[i]; solu.add(num[i]); getComboSum(num, target, sum, i+1); sum-=num[i]; solu.remove(solu.size()-1); while(i<num.length-1 && num[i]==num[i+1]) i++; //比较合适的做法 } } }