15. 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)
【思路】先整体排序，固定第一个数，第二个数从左往右，第三个数从右往左，判断他们三个的和。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
        vector<int> temp;
        int len = nums.size();
        if(len<3) return result;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < len; i++)
        { 
           while((i>0)&& i < len && (nums[i]==nums[i-1]))
            i++;
           int j = i+1;
           int k = len-1;
           while(j < k)
           {
               int sum = nums[i]+nums[j]+nums[k];
               if(sum==0)
               {
                   temp.push_back(nums[i]);
                   temp.push_back(nums[j]);
                   temp.push_back(nums[k]);
                   result.push_back(temp);
                   temp.clear();
                   j++;
                   k--;
               
               while(j < k && nums[j]==nums[j-1])
                j++;
                while(j < k&& nums[k]==nums[k+1])
                k--;
               }
                else if(sum < 0)
                {
                    j ++;
                    //skip same j
                    while(j < k && nums[j] == nums[j-1])
                        j ++;
                }
                else
                {
                    k --;
                    //skip same k
                    while(k > j && nums[k] == nums[k+1])
                        k --;
           }
        }
        }
        return result;
    }
};