leetcode—3sum

1.题目描述

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)

2.解法分析

之前做过3sum closest的题目，很显然，那里的思路应用到这里是绝对可行的， 但是这个题目我觉得可以用hashmap来做，结果就写了个基于hash的程序，可是结果总是差点，检查了好半天没检查出来，先记录一下

class Solution { public:   vector<vector<int> > threeSum(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<vector<int> > result; int numSize=num.size(); if(numSize<2)return result; sort(num.begin(),num.end()); unordered_multiset<int> myHash; for(int i =0;i<num.size();++i) { myHash.insert(num[i]); } int thirdNum=0; vector<int>cur; cur.assign(3,1); for(int i=0;i<num.size()-2;++i) { if(i>0&&num[i-1]==num[i])break; if(num[i]>0)break; if(num[i]!=num[i+1]&&myHash.count(num[i])>0)myHash.erase(num[i]); for(int j=i+1;j<num.size()-1;++j) { thirdNum=0-num[i]-num[j]; if(thirdNum<num[j])break; if(num[j+1]!=num[j])myHash.erase(num[j]); if(myHash.count(thirdNum)>0) { if(cur[0]!=num[i]||cur[1]!=num[j]||cur[2]!=thirdNum) { cur[0]=num[i];cur[1]=num[j];cur[2]=thirdNum; result.push_back(cur); } } } } return result; }   };