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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:
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]
本题和Combination Sum 非常类似，也是从一组数中找到其和为指定值的所有组合。但是本题的特殊之处在于每个给出的候选数只能用一次，且组合不能重复。
代码如下：
public class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { Arrays.sort(candidates); solve(candidates, 0, target, new LinkedList<Integer>()); return ans; } private List<List<Integer>> ans=new LinkedList<>(); private void solve(int[] candidates,int start,int target,LinkedList<Integer> current){ if (target==0){ ans.add(copy(current)); return; } if (start>=candidates.length){ return; } if (target<candidates[start]){ return; } else { for (int iterator=start;iterator<candidates.length;iterator++){ if (candidates[iterator]>target){ break; } //如果该节点和上一个节点值相同，那么它的所有组合必然包括在上一个节点的所有组合里，必须跳过 if (iterator>start&&candidates[iterator]==candidates[iterator-1]){ continue; } current.add(candidates[iterator]); solve(candidates, iterator+1, target - candidates[iterator], current); //每使用过iterator，就要+1，使得每个数只用一次 current.pollLast(); } return; } } private LinkedList<Integer> copy(LinkedList<Integer> source){ LinkedList<Integer> dest=new LinkedList<>(); for (int i:source){ dest.add(i); } return dest; } }
Combination Sum II 组合数求和之2-Leetcode
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原文地址：http://www.cnblogs.com/bethunebtj/p/4825307.html
