【Leetcode】之 Combination Sum II

一.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.
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]

二.我的解题思路

这道题是上一道题的变种，程序变动的地方也很少。测试通过的程序如下：

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {         sort(candidates.begin(), candidates.end());        vector<int> curr;        vector<vector<int>> res;        int len = candidates.size();       for(int i=0;i<len;i++){           if(i>0&&candidates[i]==candidates[i-1])                continue;     //防止找到重复的解           int curr_min=candidates[i];           int curr_target=target-curr_min;            curr.push_back(curr_min);           judge(candidates,i,curr_target,curr,res);           curr.pop_back();       }       return res;    }    void judge(vector<int>& candidates, int min_idx,int target,vector<int>&curr,vector<vector<int>>& res ){     int len = candidates.size();    if (target == 0)    {        //curr.push_back(candidates[min_idx]);        res.push_back(curr);    //  curr.pop_back();    }    if (target>=candidates[min_idx]){        for (int i = min_idx+1; i<len; i++){                if(i>min_idx+1&&candidates[i]==candidates[i-1]) //防止找到重复的解                continue;             int curr_min = candidates[i];            int curr_target = target - curr_min;            curr.push_back(curr_min);            judge(candidates, i, curr_target, curr, res);            curr.pop_back();        }    }  } };

本题的易错点在于解集不得包含重复的解。而按照上一题的解法，会包含重复解，所以需要在程序中加上标注的两处判断语句。