C++常见的stl容器与相关操作 示例解析

目录

• sort排序
• vector
• map
• unordered_map
• set
• queue
• stack
• 创建容器时指定排序规则

sort排序

针对含有迭代器的容器，可以用#include<algorithm>中的sort函数进行排序。

默认排序是从小到大，可以自己写仿函数，也可以用greater<int>()或者less<int>()。

#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
bool compfunc(const int &a, const int &b)
{
    return a > b;
}
struct compstru
{
    bool operator()(int a, int b)
    {
        return a > b;
    }
};
int main()
{
    vector<int> myVec1 = {1, 4, 9, 2};
    sort(myVec1.begin(), myVec1.end(), compstru());
    sort(myVec1.begin(), myVec1.end(), compfunc);
    sort(myVec1.begin(), myVec1.end());
    sort(myVec1.begin(), myVec1.end(), less<int>());
    sort(myVec1.begin(), myVec1.end(), greater<int>());
}

vector

void printVec(const vector<int> &vec)
{
    for (auto i : vec)
    {
        cout << i << " ";
    }
    cout << endl;
}
int main()
{
    vector<int> myVec1 = {1, 4, 9, 2};
    reverse(myVec1.begin(), myVec1.end()); // 反转vector
    printVec(myVec1);
    if(find(myVec1.begin(), myVec1.end(), 4)!=myVec1.end()){ //查找4是不是在vector里面
        cout<<"找到了"<<endl;
    }
    int row = 5;
    int col = 10;
    vector<vector<int>> myVec2(row, vector<int>(col, 0)); // 初始化全0的二维数组
    myVec1.push_back(10);
    myVec1.pop_back();
}

map

红黑树实现是有序容器，按照key值从小到大排序，插入pair<type1,type2>(data1,data2)

void printMap(const map<int, char> &myMap)
{
    for (auto it : myMap)
    {
        cout << it.second << " ";
    }
    cout << endl;
}
int main()
{
    map<int, char> myMap = {{3, 'c'}, {2, 'b'}, {1, 'a'}};
    printMap(myMap); // a b c
    myMap.insert({4, 'd'});
    printMap(myMap);                       // a b c d
    myMap.insert(pair<int, char>(0, 'e')); // e a b c d
    printMap(myMap);
    myMap[6] = 'g';                 //有覆盖数据的危险
    printMap(myMap);                // e a b c d g
    cout << myMap.count(3) << endl; // map的查找，返回1或0
    auto it = myMap.find(3);
    cout << it->second << endl; // c
    if (myMap.find(3) != myMap.end())
    {
        cout << myMap[3] << endl; // c
    }
}

unordered_map

无序容器，操作和map类似

新增元素传送门

有insert和emplace

void printMap(const unordered_map<int, char> &myMap)
{
    for (auto it : myMap)
    {
        cout << it.second << " ";
    }
    cout << endl;
}
int main()
{
    unordered_map<int,char> myUnorderMap={{1,'a'},{2,'b'},{3,'d'}};
    printMap(myUnorderMap);
    myUnorderMap.emplace(4,'e');
    myUnorderMap.insert({5,'d'});
    printMap(myUnorderMap);
    for(auto it=myUnorderMap.begin();it!=myUnorderMap.end();){
        if(it->first==1){
            myUnorderMap.erase(it++); //删除要使用it++，避免迭代器实效
        }
        else{
            it++; //迭代器递增
        }
    }
    printMap(myUnorderMap);
}

set

有序容器，会自动排序，默认从小到大

void printSet(const set<int> &mySet)
{
    for (auto it = mySet.begin(); it != mySet.end(); it++)
    {
        cout << *it << " ";
    }
    cout << endl;
}
int main()
{
    set<int> mySet = {1, 1, 2, 3};
    printSet(mySet);
    mySet.insert(0);
    printSet(mySet);
}

queue

队列

int main()
{
    queue<int> myQue; //默认使用deque作容器适配器
    for (int i = 0; i < 3; i++)
    {
        myQue.push(i);
    }
    int top = myQue.front();
    myQue.pop();
}

stack

栈

int main()
{
    stack<int> mystack;
    for(int i=0;i<4;i++){
        mystack.push(i);
    }
    int top = mystack.top();
    mystack.pop();
}

创建容器时指定排序规则

针对有序容器使用，如map，set。vector和unordered_map则不行

struct compstru
{
    bool operator()(int a, int b)
    {
        return a > b;
    }
};
template <class T1, class T2>
void printSet(const set<T1, T2> &mySet)
{
    for (auto it = mySet.begin(); it != mySet.end(); it++)
    {
        cout << *it << " ";
    }
    cout << endl;
}
template <class T1, class T2, class T3>
void printMap(const map<T1, T2, T3> &mySet)
{
    for (auto it = mySet.begin(); it != mySet.end(); it++)
    {
        cout << it->second << " ";
    }
    cout << endl;
}
int main()
{
    set<int, compstru> mySet = {1, 1, 2, 3};
    printSet<int, compstru>(mySet); // 3 2 1
    map<int, char, compstru> myMap = {{1, 'c'}, {2, 'b'}, {3, 'd'}};
    printMap<int, char, compstru>(myMap); // d b c
}