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[Leetcode] Combination Sum II
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]
Solution:
1 public class Solution { 2 public List<List<Integer>> combinationSum2(int[] candidates, int target) { 3 List<List<Integer>> result=new ArrayList<List<Integer>>(); 4 List<Integer> al=new ArrayList<Integer>(); 5 Arrays.sort(candidates); 6 dfs(result,al,target,candidates,0); 7 return result; 8 } 9 10 private void dfs(List<List<Integer>> result, List<Integer> al, int target, int[] candidates, int position) {11 // TODO Auto-generated method stub12 if(target<0)13 return;14 if(target==0){15 result.add(new ArrayList<Integer>(al));16 return;17 }18 for(int i=position;i<candidates.length;++i){19 al.add(candidates[i]);20 dfs(result, al, target-candidates[i], candidates, i+1);21 al.remove(al.size()-1);22 while((i+1<candidates.length)&&candidates[i]==candidates[i+1])23 ++i;24 } 25 }26 }
这道题和Combination Sum那道题很像，区别就是在第20行和第22行，
如果candidates[i]==candidates[i+1]，就直接跳过candidates[i+1]这个选项了，因为不能有重复的solution嘛。
[Leetcode] Combination Sum II
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