The Solution to LeetCode 15 3Sum改进版

Question:

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: The solution set must not contain duplicate triplets.

上一篇的这道题的解答有点问题，结果中出现了冗余，本来想在上篇的代码的基础上做出改进，直接写出去掉冗余结果的代码，但是没 试成功，网上看了看别人对这道题的思路，有一点点启发，根据所给的例子，我的思路如下图：

Answer:

class Solution {public:    vector<vector<int> >*r;    vector<vector<int>> threeSum(vector<int>& nums) {                r= new vector<vector<int>>();        sort(nums.begin(), nums.end());        for(int i = 0; i<nums.size(); i++)        {            //跳过相同的i            while(i>0 && i<nums.size ()&& nums[i] == nums[i-1])                  i++;                              int j = i + 1;            int p = nums.size()- 1;                        while(j < p)            {                if(nums[i] + nums[j] + nums[p] == 0)                {                        vector<int> tmp;                            tmp.push_back(nums[i]);                            tmp.push_back(nums[j]);                            tmp.push_back(nums[p]);                            r->push_back(tmp);                    j ++;                    p --;                    //跳过相同的j                    while(j < p && nums[j] == nums[j-1])                        j ++;                    //跳过相同的p                    while(p> j && nums[p] == nums[p+1])                        p --;                }                else if(nums[i] + nums[j] + nums[p] < 0)                {                    j ++;                    //跳过相同的j                    while(j < p && nums[j] == nums[j-1])                        j ++;                }                else                {                    p --;                    //跳过相同的p                    while(p > j && nums[p] == nums[p+1])                        p --;                }            }        }        return *r;    }  }; 

Run Code Result:

Your input

[-1,0,1,2,-1,-4]

Your answer

[[-1,-1,2],[-1,0,1]]

Expected answer

[[-1,-1,2],[-1,0,1]]