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;

    }

};