1. map[key]

通过键直接查找，如果存在就返回对应的值，如果不存在则返回0

	map<char, int>map1;
	map1['a'] = 1;
	map1['b'] = 2;

	cout << map1['a'] << endl; // 返回1
	cout << map1['c'] << endl; // 返回0

2. map.find(key)

返回key对应的迭代器，如果不存在则返回map.end()，时间复杂度为O(logN)

	if (map1.find('d') == map1.end())
		cout << "NOT FONUND" << endl;
	cout << map1.find('a')->second << endl; // 输出1

3. map.count(key)

如果key存在就返回1，如果不存在则返回0。

	cout << "map.count():" << endl;
	cout << map1.count('b') << endl; // 返回1
	cout << map1.count('d') << endl; // 返回0

完整测试代码：

#include<bits/stdc++.h>
using namespace std;

int main() {
	map<char, int>map1;
	map1['a'] = 1;
	map1['b'] = 2;

	cout << map1['a'] << endl; // 返回1
	cout << map1['c'] << endl; // 返回0

	cout << "map.find():" << endl;
	if (map1.find('d') == map1.end())
		cout << "NOT FONUND" << endl;
	cout << map1.find('a')->second << endl; // 输出1

	cout << "map.count():" << endl;
	cout << map1.count('b') << endl; // 返回1
	cout << map1.count('d') << endl; // 返回0

	return 1;
}

发现一个有趣的问题：

输出一个不存在的key的map映射值时，会把这个值存到map1里面，0为对应的value。
cout<<map[不存在的key];

#include<bits/stdc++.h>
using namespace std;

int main() {
	map<char, int>map1;
	map1['a'] = 1;
	map1['b'] = 2;

	cout << map1['a'] << endl; // 返回1
	cout << map1['c'] << endl; // 这里相当于存入了['c',0]到map1中

	cout << "map.find():" << endl;
	if (map1.find('c') == map1.end())
		cout << "NOT FONUND" << endl;
	cout << map1.find('c')->second << endl; // 返回0

	cout << "map.count():" << endl;
	cout << map1.count('b') << endl; // 返回1
	cout << map1.count('c') << endl; // 'c'存在所以返回1

	return 1;
}