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

求出给定整数序列中三个之和为0的不重复组合序列！

我采用了暴力解决办法，三层循环，但是Status: Time Limit Exceeded~~~
还没想出更高效的代码。

Time Limit Exceeded代码

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> vv;

        int size = nums.size();

        if (size < 3)
            return vv;

        for (int i = 0; i < size; i++)
        {
            for (int j = i + 1; j < size; j++)
            {
                for (int k = j + 1; k < size; k++)
                {
                    if (nums[i] + nums[j] + nums[k] == 0)
                    {
                        int arr[3] = { nums[i], nums[j], nums[k] };
                        //将arr中的元素从小到大排序
                        sort(arr);

                        vector<int> v = { arr, arr + 3 };
                        if (!find(vv , v))
                            vv.push_back(v);
                    }
                    else{
                        continue;
                    }
                }//for
            }//for
        }//for

        return vv;
    }

    bool find(vector<vector<int>> &vv, vector<int> &v)
    {
        vector<vector<int>>::iterator iter;

        for (iter = vv.begin(); iter != vv.end(); iter++)
        {       
            if ((*iter)[0] == v[0] && (*iter)[1] == v[1] && (*iter)[2] == v[2])
            {
                return true;
            }                       
        }
        return false;
    }

    void sort(int *a)
    {
        for (int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 3 - i - 1; j++)
            {
                if (a[j] > a[j + 1])
                {
                    int t = a[j];
                    a[j] = a[j + 1];
                    a[j + 1] = t;
                }
            }
        }
    }
};

AC代码

待补充~~~