Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.Note:All numbers (including target) will be positive integers.The solution set must not contain duplicate combinations.Example 1:

Input: candidates = [2,3,6,7], target = 7,A solution set is:[  [7],  [2,2,3]]

Example 2:

Input: candidates = [2,3,5], target = 8,A solution set is:[  [2,2,2,2],  [2,3,3],  [3,5]]

难度：medium

题目：给定一无重复元素的集合和一指定数，找出所有由数组内元素之和为该数的序列。每个元素都可以被无限次使用。

注意：所以元素都为正整数包括给定的整数。答案不允许有重复的组合。

思路：递归

Runtime: 9 ms, faster than 81.90% of Java online submissions for Combination Sum.

Memory Usage: 29.6 MB, less than 16.71% of Java online submissions for Combination Sum.

class Solution {    public List
    
     
      > combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        List
      
       
        > result = new ArrayList();        combinationSum(candidates, 0, target, 0, new Stack
        
         (), result);                return result;    }        public void combinationSum(int[] cs, int idx, int target, int sum, Stack
         
           stack, List
          
           
            > result) { if (sum == target) { result.add(new ArrayList(stack)); return; } for (int i = idx; i < cs.length; i++) { if (sum + cs[i] <= target) { stack.push(cs[i]); combinationSum(cs, i, target, sum + cs[i], stack, result); stack.pop(); } if (sum + cs[i] >= target) break; } }}