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]]