40. Combination Sum II

题目：

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.

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.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

链接： http://leetcode.com/problems/combination-sum-ii/

4/14/2017

18ms, 92%

出现的问题

1. 第6行排序不要遗忘

2. 第17行最开始的判断为什么是i > pos？我直接想想不出来，但是可以尝试着解释[1,1,2,5,6,7,10]中[1,1,6]和[1,7]第一组为什么可以有2个1？因为这两个1不是在同一层加进去的，所以可以视为不同的元素，而[1,7]不能重复是因为当时2个1是在同一层，pos值是一样的。

3. 时间复杂度？

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> ret = new ArrayList<>();
        if (candidates.length == 0) return ret;
        List<Integer> temp = new ArrayList<Integer>();
        Arrays.sort(candidates);
        enumerate(candidates, target, ret, temp, 0);
        return ret;
    }
    
    private void enumerate(int[] candidates, int target, List<List<Integer>> ret, List<Integer> temp, int pos) {
        if (target == 0) {
            ret.add(new ArrayList<Integer>(temp));
            return;
        }
        for (int i = pos; i < candidates.length; i++) {
            if (i > pos && candidates[i] == candidates[i - 1]) continue;
            if (candidates[i] > target) break;
            temp.add(candidates[i]);
            enumerate(candidates, target - candidates[i], ret, temp, i + 1);
            temp.remove(temp.size() - 1);
        }
    }
}

更多讨论：

https://discuss.leetcode.com/category/48/combination-sum-ii