Map container

#include<map>
using std::map;

1.map The definition of

map<typename1,typename2> name;

typename1 As the key (key),keyname2 Value (value)

2.map Access to elements within a container

（1） The subscript access

Note that a key can only correspond to one value

map<char,int> mp;
mp['a']=100;
mp['a']=120;
cout<<mp['a'];
// Output 120

（2） Iterators access

Iterator definitions and others STL identical

map<typename1,typename2>::iterator it;

Use and others STL There's a difference , Because iterators need to be able to access keys and values at the same time , Use ->first and ->second Access keys and values respectively

Be careful ：map According to key Automatically sort the size of , Be similar to set

#include<bits/stdc++.h>
using namespace std;
signed main(void)
{
    map<char,int> mp;
    mp['c']=1;
    mp['b']=2;
    mp['a']=3;
    for(map<char,int>::iterator it=mp.begin();it!=mp.end();it++)
    {
        printf("%c %d\n",it->first,it->second);
    }
    return 0;
}
/*
 The output is ( Automatic sorting )
a 3
b 2
c 1
*/

3.map Common function analysis

（1）find()

find(key) Return key is key The iterator of the map of , The time complexity is O(logn),n by map Number of mappings in

#include<bits/stdc++.h>
using namespace std;
signed main()
{
    map<char,int> mp;
    mp['b']=2;
    mp['c']=1;
    mp['a']=3;
    map<char,int>::iterator it=mp.find('b');
    printf("%c %d\n",it->first,it->second);
    return 0;
}

（2）erase()

1.erase(it) it Iterator for the element to be deleted , Time complexity O(1)

2.erase(key) key Is the key of the element to be deleted , The time complexity is O(logn),n by map Number of internal elements

3.erase(first,last) Left closed right away ,[first,last), The time complexity is last-first

（3）size()

Return mapping logarithm ,O(1)

（4）clear()

Empty elements ,O(n),n Is the number of elements

4.map Common use

multimap（ A key corresponds to multiple values ）、unordered_map（ Don't order ）