Leetcode 40 Combination Sum II

triplebee

发布于 2018-01-12 15:05:57

4430

发布于 2018-01-12 15:05:57

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]
]

与39类似，只是注意数字不能重复使用

https://cloud.tencent.com/developer/article/1019276

因而开始不用去重，搜索的过程中要注意避免重复搜索相同分支的情况

class Solution {
public:
    void dfs(vector<int> candidates,int target,vector<vector<int>>& result,vector<int> path)  
    {  
        if(target==0)  
        {  
            result.push_back(path);      
            return ;  
        } 
        int t;
        for(int i=candidates.size()-1;i>=0;i--)  
        {  
            if(candidates[i]==t) 
            {
                candidates.pop_back();
                continue;
            }
            t=candidates[i];
            candidates.pop_back();
            if(t>target) continue;  
            path.push_back(t);  
            dfs(candidates,target-t,result,path);  
            path.pop_back();  
        }  
    }  
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target)   
    {  
        vector<vector<int>> result;  
        vector<int> path;  
        sort(candidates.begin(),candidates.end());  
        dfs(candidates,target,result,path);  
        return result;  
    } 
};

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原始发表：2016-09-06 ，如有侵权请联系 cloudcommunity@tencent.com 删除

combinations