3Sum

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)

解题思路：
（1）将数组递增排序
（2）依次枚举第一个加数
（3）设置两个指针分别指向第二个和第三个加数，第二个加数从前往后，第三个加数从后往前
（4）元素去重的方法

class Solution {
public:

vector<vector<int>> threeSum(vector<int>& nums){    sort(nums.begin(), nums.end());//对数组元素进行递增排序    vector<vector<int>> res;    int numsSize = nums.size();    for (int i = 0; i < numsSize; i++)    {        //对元素nums[i]去重        if (i > 0 && nums[i] == nums[i-1])            continue;        int j = i+1;        int k = numsSize-1;        while (j < k)        {            int sum = nums[i]+nums[j]+nums[k];            if (sum < 0)            {                j++;                //对元素nums[j]去重                while (nums[j] == nums[j-1] && j < k)                {                    j++;                }            }            else if (sum > 0)            {                k--;                //对元素nums[k]去重                while (nums[k] == nums[k+1] && j < k)                {                    k--;                }            }            else            {                vector<int> temp(3);                temp[0] = nums[i];                temp[1] = nums[j++];                temp[2] = nums[k--];                res.push_back(temp);                //对元素nums[j]去重                while (nums[j] == nums[j-1] && j < k)                {                    j++;                }                //对元素nums[k]去重                while (nums[k] == nums[k+1] && j < k)                {                    k--;                }            }        }    }    return res;    }

};