【LeetCode】040. 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.
• 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]
]

题解：

Solution 1 (TLE)

class Solution {
public:
    void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, int sum, vector<int>& visited) {
        if(sum == target) {
            vector<int>* tmp = new vector<int>;
            *tmp = v;
            sort((*tmp).begin(), (*tmp).end());
            if(find(vv.begin(), vv.end(), *tmp) == vv.end())
                vv.push_back(*tmp);
            delete tmp;
            return;
        }
        if(sum > target) return;
        for(int i=0; i<candidates.size(); ++i) {
            if(visited[i] != 0) continue;
            v.push_back(candidates[i]);
            visited[i] = 1;
            dfs(vv, v, candidates, target, sum+candidates[i], visited);
            v.pop_back();
            visited[i] = 0;
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> vv;
        vector<int> v;
        vector<int> visited(candidates.size(),0);
        dfs(vv, v, candidates, target, 0, visited);
        return vv;        
    }
};

　　Solution 1 中即使i从level开始遍历，也无法accepted，加入stop后才勉强通过，即Solution 2 

Solution 2 (almost TLE but not 579ms)

class Solution {
public:
    void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, int sum, vector<int>& visited, int stop,int level) {
        if(sum == target) {
            vector<int>* tmp = new vector<int>;
            *tmp = v;
            sort((*tmp).begin(), (*tmp).end());
            if(find(vv.begin(), vv.end(), *tmp) == vv.end())
                vv.push_back(*tmp);
            delete tmp;
            return;
        }
        if(sum > target) {stop = 1;return;}
        for(int i=level; i<candidates.size(); ++i) {
            if(visited[i] != 0) continue;
            v.push_back(candidates[i]);
            visited[i] = 1;
            dfs(vv, v, candidates, target, sum+candidates[i], visited, stop, level+1);
            if(stop == 1) {stop = 0;break;}
            v.pop_back();
            visited[i] = 0;
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> vv;
        vector<int> v;
        vector<int> visited(candidates.size(),0);
        sort(candidates.begin(), candidates.end());
        dfs(vv, v, candidates, target, 0, visited, 0, 0);
        return vv;        
    }
};

Solution 3 ()

class Solution {
public:
    void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, vector<int>& visited, int stop,int level) {
        if(target == 0) {
            vector<int>* tmp = new vector<int>;
            *tmp = v;
            sort((*tmp).begin(), (*tmp).end());
            if(find(vv.begin(), vv.end(), *tmp) == vv.end())
                vv.push_back(*tmp);
            delete tmp;
            return;
        }
        if(target<0) {stop = 1;return;}
        for(int i=level; i<candidates.size(); ++i) {
            if(visited[i] != 0) continue;
            v.push_back(candidates[i]);
            visited[i] = 1;
            dfs(vv, v, candidates, target-candidates[i], visited, stop,level+1);
            if(stop == 1) {stop = 0;break;}
            v.pop_back();
            visited[i] = 0;
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> vv;
        vector<int> v;
        vector<int> visited(candidates.size(),0);
        sort(candidates.begin(), candidates.end());
        dfs(vv, v, candidates, target, visited, 0,0);
        return vv;        
    }
};

Solution 4