# Combination Sum II ### Source - leetcode: [Combination Sum II | LeetCode OJ](https://leetcode.com/problems/combination-sum-ii/) - lintcode: [(153) Combination Sum II](http://www.lintcode.com/en/problem/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. Have you met this question in a real interview? Yes Example For example, given candidate set 10,1,6,7,2,1,5 and target 8, A solution set is: [1,7] [1,2,5] [2,6] [1,1,6] 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. ~~~ ### 题解 和 [Unique Subsets](http://algorithm.yuanbin.me/zh-cn/exhaustive_search/unique_subsets.html) 非常类似。在 [Combination Sum](http://algorithm.yuanbin.me/zh-cn/exhaustive_search/combination_sum.html) 的基础上改改就好了。 ### Java ~~~ public class Solution { /** * @param num: Given the candidate numbers * @param target: Given the target number * @return: All the combinations that sum to target */ public List<List<Integer>> combinationSum2(int[] num, int target) { List<List<Integer>> result = new ArrayList<List<Integer>>(); List<Integer> list = new ArrayList<Integer>(); if (num == null) return result; Arrays.sort(num); helper(num, 0, target, list, result); return result; } private void helper(int[] nums, int pos, int gap, List<Integer> list, List<List<Integer>> result) { if (gap == 0) { result.add(new ArrayList<Integer>(list)); return; } for (int i = pos; i < nums.length; i++) { // ensure only the first same num is chosen, remove duplicate list if (i != pos && nums[i] == nums[i - 1]) { continue; } // cut invalid num if (gap < nums[i]) { return; } list.add(nums[i]); // i + 1 ==> only be used once helper(nums, i + 1, gap - nums[i], list, result); list.remove(list.size() - 1); } } } ~~~ ### 源码分析 这里去重的方法继承了 Unique Subsets 中的做法，当然也可以新建一变量 `prev`，由于这里每个数最多只能使用一次，故递归时索引变量传`i + 1`. ### 复杂度分析 时间复杂度 O(n)O(n)O(n), 空间复杂度 O(n)O(n)O(n).