Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.Each number in candidates may only be used once in the combination.Note:All numbers (including target) will be positive integers.The solution set must not contain duplicate combinations.Example 1:Input: candidates = [10,1,2,7,6,1,5], target = 8,A solution set is:[ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6]]Example 2:Input: candidates = [2,5,2,1,2], target = 5,A solution set is:[ [1,2,2], [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。每个元素在组合中仅可被使用一次。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: 10 ms, faster than 84.44% of Java online submissions for Combination Sum II.Memory Usage: 31.5 MB, less than 6.01% of Java online submissions for Combination Sum II.class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { Arrays.sort(candidates); List<List<Integer>> result = new ArrayList(); combinationSum(candidates, 0, target, 0, new Stack<>(), result); return result; } private void combinationSum(int[] cs, int idx, int target, int s, Stack<Integer> stack, List<List<Integer>> r) { if (s == target) { r.add(new ArrayList(stack)); return; } for (int i = idx; i < cs.length; i++) { if (s + cs[i] <= target && (i == idx || cs[i - 1] != cs[i])) { stack.push(cs[i]); combinationSum(cs, i + 1, target, s + cs[i], stack, r); stack.pop(); } } }}