Leetcode--Combination Sum II

最新推荐文章于 2021-03-18 11:37:34 发布

yskyj

最新推荐文章于 2021-03-18 11:37:34 发布

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分类专栏： Leetcode 文章标签： leetcode Combination Sum II

本文链接：https://blog.youkuaiyun.com/yskyj/article/details/41076925

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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]

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Array Backtracking

Have you met this question in a real interview?

思路：还是沿着问题1的解题思路，需要额外注意的是重复问题，如：[1,1,1] target=1,得到的结果不能是三个[1]，那么如果消除重复呢，最简单的方法就是用set容器

class Solution {
public:
    vector<vector<int>> res;
    set<vector<int>> result;
    void backtrace(vector<int> v,vector<int>&num,int target,int pos,int sum)
    {
        if(sum==target)
            result.insert(v);//res.push_back(v);
        else if(sum>target||pos>=num.size())
            return ;
        else if(sum<target){
            for(int i=pos;i<num.size();i++)
            {
                vector<int> tmp(v);
                tmp.push_back(num[i]);
                backtrace(tmp,num,target,i+1,sum+num[i]);
            }
        }
    }
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        if(num.size()<1)
            return res;
            
        sort(num.begin(),num.end());
        vector<int> v;
        backtrace(v,num,target,0,0);
        res.assign(result.begin(),result.end());
        return res;
    }
};
Submission Result: Time Limit Exceeded

Last executed input:	[6,23,31,29,16,25,14,7,14,17,9,24,24,13,26,14,30,26,24,16,20,8,27,25,16,14,18,12,7,27,16,9,7,13,25,19,33,26,21,5,9,18,16,20,26,7,21,21,21,32,24,28,27,17,18,9,18,23,16,20,15,19,16,9], 24

但是得到TLE，这是怎么回事呢？

分析可能的原因在于backtrace中连续声明并初始化tmp对象，带来了额外的时间开销。

优化该段代码：

class Solution {
public:
    vector<vector<int>> res;
    set<vector<int>> result;
    void backtrace(vector<int> &v,vector<int>&num,int target,int pos,int sum)
    {
        if(sum==target)
            result.insert(v);//res.push_back(v);
        else if(sum>target||pos>=num.size())
            return ;
        else if(sum<target){
            for(int i=pos;i<num.size();i++)
            {
                
                v.push_back(num[i]);
                backtrace(v,num,target,i+1,sum+num[i]);
                v.pop_back();
            }
        }
    }
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        if(num.size()<1)
            return res;
            
        sort(num.begin(),num.end());
        vector<int> v;
        backtrace(v,num,target,0,0);
        res.assign(result.begin(),result.end());
        return res;
    }
};
Submission Result: Accepted