LeetCode - 1. Two Sum
1. Two Sum
Problem's Link
Mean:
给定一个数组nums和一个数target,求:id1和id2,id1和id2为数组nums两个不同的下标,使得nums[id1]+nums[id2]=target.
注意:nums中元素可重.
analyse:
如果nums中没有重复元素,那么可以用map做.
由于有重复元素,需要用multimap.
注意:使用map时,需要用count(key)来检测是否存在key值,否则会出现错误.
Time complexity: O(N*logN)
view code
/** * ----------------------------------------------------------------- * Copyright (c) 2016 crazyacking.All rights reserved. * ----------------------------------------------------------------- * Author: crazyacking * Date : 2015-01-29-14.24 */ #include <queue> #include <cstdio> #include <set> #include <string> #include <stack> #include <cmath> #include <climits> #include <map> #include <cstdlib> #include <iostream> #include <vector> #include <algorithm> #include <cstring> using namespace std; typedef long long(LL); typedef unsigned long long(ULL); const double eps(1e-8); class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { int cnt=0; vector<int> ans; multimap<int,int> mp; for(auto p:nums) mp.insert(make_pair(p,cnt++)); for(auto p:mp) { if(mp.count(target-p.first)>0) { multimap<int,int>::iterator it1,it2; if((p.first==target-p.first)&&(mp.count(p.first)==2)) { it1=mp.find(p.first); ans.push_back((*it1).second+1); mp.erase(it1); it2=mp.find(p.first); ans.push_back((*it2).second+1); break; } it1=mp.find(p.first); it2=mp.find(target-p.first); ans.push_back((*it1).second+1); ans.push_back((*it2).second+1); break; } } sort(ans.begin(),ans.end()); return ans; } }; int main() { int n; while(cin>>n) { vector<int> ve; for(int i=0,tmp;i<n;++i) { cin>>tmp; ve.push_back(tmp); } int target; cin>>target; Solution a; vector<int> ans=a.twoSum(ve,target); for(auto p : ans) cout<<p<<endl; } return 0; }