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]
]

这道题与39题类似，不过这里不允许使用相同元素两次。

#include <vector>using namespace std;class CombinationSumII{public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target);    void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int begin);};

#include <algorithm>#include "CombinationSumII.h"vector<vector<int>> CombinationSumII::combinationSum2(vector<int>& candidates, int target){    sort(candidates.begin(), candidates.end());    vector<std::vector<int> > res;    vector<int> combination;    //开始迭代    combinationSum(candidates, target, res, combination, 0);    return res;}void CombinationSumII::combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {    if (!target) {        res.push_back(combination);        return;    }    for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {        //这句是与上一道题的区别点        if (i > begin && candidates[i] == candidates[i - 1]) continue;        combination.push_back(candidates[i]);        //每次查找的目标元素不同        combinationSum(candidates, target - candidates[i], res, combination, i);        //这步保证遍历完所有子集        combination.pop_back();    }}