#include <iostream>
#include <vector>#include <stdio.h>#include <map>using namespace std;class Solution {public: vector<int> twoSum(vector<int> &numbers, int target) { vector<int> ret(2,0); }};int main(){ multimap<int,char*> m; //multimap的插入只能用insert()不能用数组 //muiltmap 初始化操作和insert操作 m.insert(pair<int,char*>(1,"apple")); m.insert(pair<int,char*>(1,"pear"));//apple和pear的价钱完全有可能是一样的 m.insert(pair<int,char*>(2,"banana")); //m[1]="orange"; multimap没有下标运算 //multimap的遍历只能用迭代器方式不能用数组 cout<<"***************************************"<<endl; multimap<int,char*>::iterator i,iend; iend=m.end(); for(i=m.begin();i!=iend;i++) { cout<<(*i).second<<"的价钱是"<<(*i).first<<"元/斤\n"; } cout<<"***************************************"<<endl; //元素的反相遍历 multimap<int,char*>::reverse_iterator j,jend; jend=m.rend(); for(j=m.rbegin();j!=jend;j++) { cout<<(*j).second<<"的价钱是"<<(*j).first<<"元/斤\n"; } cout<<"***************************************"<<endl; //元素的搜索find(),pair<iterator,iterator>equal_range(const key_type &k)const //和multiset的用法一样 multimap<int,char*>::iterator s; s=m.find(1);//find()只要找到一个就行了，然后立即返回。 cout<<(*s).second<<" "<<(*s).first<<endl; cout<<"键值等于1的元素个数是："<<m.count(1)<<endl; cout<<"***************************************"<<endl; //输出所以匹配项 pair<multimap<int,char*>::iterator,multimap<int,char*>::iterator> range; range = m.equal_range(1); for(multimap<int,char*>::iterator it = range.first;it!=range.second;it++) cout<<it->second<<endl; cout<<"***************************************"<<endl; //删除 erase(),clear() m.erase(1); for(i=m.begin();i!=iend;i++) { cout<<(*i).second<<"的价钱是"<<(*i).first<<"元/斤\n"; } return 0;}
/*Given an array of integers, find two numbers such that they add up to a specific target number.The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.You may assume that each input would have exactly one solution.*/#include <iostream>#include <vector>#include <map>#include<stdlib.h>#include <algorithm>using namespace std;int numrand(int num){ return rand()%num;}class Solution {public: vector<int> twoSum(vector<int> &numbers, int target) { vector<int> ret(2,0); multimap<int,int> num; vector<int>::iterator vit; multimap<int,int>::iterator mmit,findn,it; int temp = 0; for(vit=numbers.begin();vit!=numbers.end();vit++,temp++) { num.insert(pair<int,int>(*vit,temp)); } for(mmit=num.begin();mmit!=num.end();mmit++) { if((findn = num.find(target-(*mmit).first))!=num.end()) { if((*findn).first == (*mmit).first) { if(num.count((*mmit).first)>1) { pair<multimap<int,int>::iterator,multimap<int,int>::iterator> range; range = num.equal_range((*mmit).first); int i=0; for(it=range.first;it!=range.second;it++) { ret[i]=it->second+1; i=1; } } } else { ret[0]= mmit->second+1; ret[1]= findn->second+1; } } } if(ret[0]>ret[1]) { temp=ret[0]; ret[0]=ret[1]; ret[1]=temp; } return ret; }};int main(){ vector<int> nrand(5,0); vector<int> ret; //for(int i=0;i<5;i++) // nrand[i]=numrand(5); // for(int j=0;j<5;j++) // cout<<nrand[j]<<" "; //cout<<endl; nrand[0]=0; nrand[1]=4; nrand[2]=3; nrand[3]=0; Solution *sou = new Solution; ret=sou->twoSum(nrand,0); cout<<ret[0]<<" "<<ret[1]; return 0;}回顾：最开始用循环遍历的方法，复杂度为O(n*n)，现在改用map的find函数，find是用二叉树实现的，所以复杂度为log(n);遍历multimap，已自动按key的大小排序。
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