[leetcode] 40. Combination Sum II

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]]

解法一：

跟 Combination Sum的区别主要是使用了set，然后把recursive中的i变成了i+1。

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        if(target==0) return{};        set<vector<int>> res;        vector<int> out;        sort(candidates.begin(), candidates.end());        combinationSum2(candidates, target, 0, out, res);        return vector<vector<int>>(res.begin(), res.end());    }        void combinationSum2(vector<int>& candidates, int target, int idx, vector<int> &out, set<vector<int>>& res){        if(target<0) return;        else if (target==0) {            res.insert(out);            return;        }                for(int i = idx; i<candidates.size(); i++){            out.push_back(candidates[i]);            combinationSum2(candidates, target-candidates[i], i+1, out, res);            out.pop_back();        }    }};

解法二：

在recursion的循环中，加上if(i>idx&&num[i]==nums[i-1]) continue。也就是如果当前的数字的值如果之前考虑过的话，就不要再考虑了。

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        if(target==0) return{};        vector<vector<int>> res;        vector<int> out;        sort(candidates.begin(), candidates.end());        combinationSum2(candidates, target, 0, out, res);        return res;    }        void combinationSum2(vector<int>& candidates, int target, int idx, vector<int> &out, vector<vector<int>>& res){        if(target<0) return;        else if (target==0) {            res.push_back(out);            return;        }                for(int i = idx; i<candidates.size(); i++){            if(i>idx&&candidates[i]==candidates[i-1]) continue;            out.push_back(candidates[i]);            combinationSum2(candidates, target-candidates[i], i+1, out, res);            out.pop_back();        }    }};