LeetCode40：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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]

这道题和前面Combination Sum 很像，只是判断条件不同而已，题目的难度也没有增加，使用相同的思路即可求解。需要注意的有一点：
对于[1 ,1 ,2] ,3这种输入，需要剔除挑重复元素，所以在递归函数中需要剔除重复元素。

runtime:16ms

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<int> path;        vector<vector<int>> result;        helper(candidates,0,0,target,path,result);        return result;    }    void helper(vector<int> &nums,int pos,int base,int target,vector<int> &path,vector<vector<int>> &result)    {        if(base==target)        {            result.push_back(path);            return;        }        if(base>target)            return;        for(int i=pos;i<nums.size();i++)        {            path.push_back(nums[i]);            helper(nums,i+1,base+nums[i],target,path,result);            path.pop_back();            while(nums[i]==nums[i+1]) i++; //这一步需要将重复的i剔除掉，做关于递归的题目时假设上面的代码正确执行，消除重复在这一步做即可。        }    }};