# Combination Sum ### Source - leetcode: [Combination Sum | LeetCode OJ](https://leetcode.com/problems/combination-sum/) - lintcode: [(135) Combination Sum](http://www.lintcode.com/en/problem/combination-sum/) ~~~ Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times. For example, given candidate set 2,3,6,7 and target 7, A solution set is: [7] [2, 2, 3] Have you met this question in a real interview? Yes Example given candidate set 2,3,6,7 and target 7, A solution set is: [7] [2, 2, 3] 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 [Permutations](http://algorithm.yuanbin.me/zh-cn/exhaustive_search/permutations.html) 十分类似，区别在于剪枝函数不同。这里允许一个元素被多次使用，故递归时传入的索引值不自增，而是由 for 循环改变。 ### Java ~~~ public class Solution { /** * @param candidates: A list of integers * @param target:An integer * @return: A list of lists of integers */ public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> result = new ArrayList<List<Integer>>(); List<Integer> list = new ArrayList<Integer>(); if (candidates == null) return result; Arrays.sort(candidates); helper(candidates, 0, target, list, result); return result; } private void helper(int[] candidates, int pos, int gap, List<Integer> list, List<List<Integer>> result) { if (gap == 0) { // add new object for result result.add(new ArrayList<Integer>(list)); return; } for (int i = pos; i < candidates.length; i++) { // cut invalid candidate if (gap < candidates[i]) { return; } list.add(candidates[i]); helper(candidates, i, gap - candidates[i], list, result); list.remove(list.size() - 1); } } } ~~~ ### 源码分析 对数组首先进行排序是必须的，递归函数中本应该传入 target 作为入口参数，这里借用了 Soulmachine 的实现，使用 gap 更容易理解。注意在将临时 list 添加至 result 中时需要 new 一个新的对象。 ### 复杂度分析 按状态数进行分析，时间复杂度 O(n!)O(n!)O(n!), 使用了list 保存中间结果，空间复杂度 O(n)O(n)O(n). ### Reference - Soulmachine 的 leetcode 题解