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;

    }

};