3Sum——解题报告

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【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 settmp.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

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