40. Combination Sum II

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();
            }
        }
    }
}