LeetCode #40 - Combination Sum II

题目描述：

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.
• 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]
]

此题与LeetCode #39类似，那道题可以重复使用数组元素，这道题不能重复使用数组元素，而且数组可能有相等的元素，同时还要保证组合的唯一性。首先要对数组排序，保证相等的元素聚在一起，思路还是分两种情况进行递归：1.加入当前元素，2.不加入当前元素。为了避免重复，如果不加入当前元素，那么在它之后的所有和它相等的元素都不能加入。

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> cur;
        int i=0;
        int sum=0;
        sort(candidates.begin(),candidates.end());
        DFS(candidates, target, i, sum, result, cur);
        return result;
    }
    
    void DFS(vector<int>& candidates, int target, int i, int sum, vector<vector<int>>& result, vector<int> cur)
    {
        if((sum+candidates[i])>target||i>=candidates.size()) return;
        else if((sum+candidates[i])==target) 
        {
            cur.push_back(candidates[i]);
            result.push_back(cur);
            return;
        }
        else if((sum+candidates[i])<target)
        {
            cur.push_back(candidates[i]);
            DFS(candidates, target, i+1, sum+candidates[i], result, cur);
            cur.pop_back();
            #当发现重复元素时，全部跳过
            while(i<candidates.size()-1&&candidates[i]==candidates[i+1]) i++;
            DFS(candidates, target, i+1, sum, result, cur);
            return;
        }
    }
};