LeetCode40. 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]]

题意分析：这道题和LeetCode39题差不多，也是选择用回溯的方法去做，但是这道题是不允许重复使用的，所以就多了一点点处理的歩奏。

void isok(vector<int>& candidates, vector<vector<int>>& result, int target, int begin, vector<int> tmp) {if (target == 0) {result.push_back(tmp);}else {for (int i = begin; i < candidates.size(); i++) {if (candidates[i] > target) {break;}if (i == begin || candidates[i] != candidates[i - 1]) {tmp.push_back(candidates[i]);isok(candidates, result, target - candidates[i], i + 1, tmp);tmp.pop_back();}}}}

代码：

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