Combination Sum II - LeetCode

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] 

 

思路：和Combination Sum这道题基本思想是相同的。需要注意的不同点主要有两个：一个是因为每个数只能用一次，因此在枚举完后继续递归时要从下一位开始枚举，而不再是当前位置；另一个还是因为每个数只能用一次，因此要注意原数组中可能有重复数字的情况，因此这里在枚举过程中如果遇见了相同的数，则只枚举一次，然后跳过剩下的重复数字。

class Solution {
public:
    void assist(vector<int>& candidates, vector<vector<int> >& res, vector<int>& cur, int target, int st)
    {
        if (target == 0)
        {
            res.push_back(cur);
            return;
        }
        if (st == candidates.size()) return;
        for (int i = st, n = candidates.size(); i < n && candidates[i] <= target; i++)
        {
            cur.push_back(candidates[i]);
            assist(candidates, res, cur, target - candidates[i], i + 1);
            cur.pop_back();
            while (i < n && candidates[i] == candidates[i + 1])
                i++;
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int> > res;
        vector<int> cur;
        sort(candidates.begin(), candidates.end(), less<int>());
        assist(candidates, res, cur, target, 0);
        return res;
    }
};