40和的组合第二弹

Combination Sum II

问题描述：

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]]

测试代码：

class Solution {    void find_answer(vector<int> candidates,vector<vector<int> >& solutions,vector<int>& solution,int target,int begin)    {        if (!target) {            solutions.push_back(solution);            return;        }        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i)            if (i == begin || candidates[i] != candidates[i - 1]) {                solution.push_back(candidates[i]);                find_answer(candidates,solutions, solution, target - candidates[i],  i + 1);                solution.pop_back();            }    }public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        vector<vector<int> > solutions;        vector<int> solution;        sort(candidates.begin(),candidates.end());        find_answer(candidates,solutions,solution,target,0);        return solutions;    }};

性能：

参考答案：

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int>> res;        vector<int> path;        helper(candidates, 0, target, res, path);        return res;    }private:    void helper(vector<int>& candidates, int pos , int target, vector<vector<int>>& res, vector<int>& path) {        if (target == 0) {            res.push_back(path);            return;        }        if (pos >= candidates.size() || candidates[pos] > target) {            return;        }        int count = 1;        while (pos + count < candidates.size() && candidates[pos] == candidates[pos + count]) {            count++;        }        for (int i = 1; i <= count; i++) {            path.push_back(candidates[pos]);            helper(candidates, pos + count, target - i * candidates[pos], res, path);        }        for (int i = 1; i <= count; i++) {            path.pop_back();        }        helper(candidates, pos + count, target, res, path);        return;    }};

性能：