3Sum——解题报告

ゞ浴缸里的玫瑰 2022-08-08 06:12 213阅读 0赞

【题目】

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.  
  
Note:  
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)  
The solution set must not contain duplicate triplets.  
For example, given array S = \{-1 0 1 2 -1 -4\},  
  
A solution set is:  
(-1, 0, 1)  
(-1, -1, 2)

【分析】

我们做过这样的题目，给定一个数组和一个target，找出数组中的a+b = target。那么这道题目变成了a，b，c三个数。

其实，我们遍历每个a，然后target' = targer - a. 也就变成了找出b+c = target'。和上面的题目一样吧~

【代码】

上述解法的时间复杂度最快也只能达到O(n^2)。

需要解释的是，下面给出的代码在LeetCode提交的时候，一直是LTE。线下测试时对的，查找了答案之后发现，主体思路一样，只不过我用到的是set去重，可能这部分比较耗时。。。

vector< vector<int> > threeSum(vector<int>& nums) 
    {
        //if(nums.empty())
            //return;
            
        sort(nums.begin(), nums.end());
        set< vector<int> > res1; 
        set< vector<int> >::iterator iter;
        vector< vector<int> > res2;
    	vector<int> tmp;
        
        if(nums.size() < 3)
            return res2;
            
        for(int i = 0; i < nums.size() - 2; i++)
        {
            int current = nums[i]; 
            int diff = 0 - current; // the target
            
            int head = i + 1, tail = nums.size() - 1;
            while(head < tail)
            {
                if(head == i)  // skip the current
                    {head++; continue;}
                if(tail == i)
                    {tail--; continue;}
                
                if(nums[head] + nums[tail] == diff)
                {
    				//cout<<i<<' '<<head<<' '<<tail<<endl;
                    tmp.push_back(current); tmp.push_back(nums[head]); tmp.push_back(nums[tail]);
                    sort(tmp.begin(), tmp.end());  // sort
    				//cout<<tmp[0]<<' '<<tmp[1]<<' '<<tmp[2]<<endl;
    				//cout<<current<<' '<<nums[head]<<' '<<nums[tail]<<endl;
                    res1.insert(tmp);  // insert into the set
    				tmp.clear();
                    head++;
                    tail--;
                    continue;
                }
                else if(nums[head] + nums[tail] < diff)
                {
                    head++;
                    continue;
                }
                else
                {
                    tail--;
                    continue;
                }
            }
        }
        
        for(iter = res1.begin(); iter != res1.end(); ++iter)
            res2.push_back(*iter);
        
        return res2;
    }

下面给出一个ac的参考答案：

http://bbs.csdn.net/topics/390931100