持续更新中

1. count_if

//找有多少个满足lambda 的元素
//调用方式：count_if(iterator, iterator, lambda);
vector<int> a = {0, 2, 3, 4};
cout << count_if(a.begin(), a.end(), [](int a1)->bool{return a1 <= 2;});
//输出：2

2. count

//找有多少个等于val 的元素
//调用方式：count(iterator, iterator, val);
vector<int> a = {0, 2, 3, 4};
cout << count(a.begin(), a.end(), 2);
//输出：1

3.find

//找到第一个值为 val 的迭代器
//调用方式：find(iterator, iterator, val);
vector<int> a = {0, 1, 0, 2, 1};
int p = find(a.begin(), a.end(), 1) - a.begin(); //第一个值为 1的下标
int rp = a.rend() - find(a.rbegin(), a.rend(), 1) - 1;//最后一个值为 1 的下标
cout << "p = " << p << " rp = " << rp << '\n';
//输出：p = 1 rp = 4

4.find_if

//找到第一个满足lambda 的迭代器
//调用方式：find_if(iterator, iterator, lambda);
vector<int> a = {0, 10, 20, 11, 10};
int p = find_if(a.begin(), a.end(), [](int i){return i > 10;}) - a.begin();//找第一个值大于10的下标
int rp = a.rend() - find_if(a.rbegin(), a.rend(), [](int i){return i > 10;}) - 1;//找最后一个值大于10的下标
cout << "p = " << p << " rp = " << rp << '\n';
//输出：p = 2 rp = 3

5.rotate

rotate(L, M, R);简单解释一下, rotate函数里面要放3个迭代器，L，M，R，原本数组会按照$[L, M) + [M, R)$ 的方式排列，函数操作之后，数组会按照$[M, R) + [L, M) $的方式排列

//旋转数组
//调用方式：rotate(iterator, iterator, iterator);
#define debugvec(a) cout<<#a<<": "; for(int i = 0; i < a.size(); i++) cout << a[i] << ' '; cout << endl;
vector<int> b = {2, 4, 6, 8, 10};
for(int i = 0; i < b.size(); i++){
    auto a = b;// a 赋值为 b
    rotate(a.begin(), a.begin() + i, a.end());
    debugvec(a);//输出 a
}
cout << '\n';
for(int i = 0; i < 4; i++){
    auto a = b;// a 赋值为 b
    rotate(a.begin(), a.begin() + i, a.begin() + 4);
    debugvec(a);//输出 a
}

函数输出

a: 2 4 6 8 10 
a: 4 6 8 10 2 
a: 6 8 10 2 4 
a: 8 10 2 4 6 
a: 10 2 4 6 8 

a: 2 4 6 8 10 
a: 4 6 8 2 10 
a: 6 8 2 4 10 
a: 8 2 4 6 10 

6.unique

#include <algorithm>
#include <iostream>
#include <vector>

int main()
{
    // a vector containing several duplicate elements
    std::vector<int> v{1,2,1,1,3,3,3,4,5,4};
    auto print = [&] (int id) {
        std::cout << "@" << id << ": ";
        for (int i : v)
            std::cout << i << ' ';
        std::cout << '\n';
    };
    print(1);

    // remove consecutive (adjacent) duplicates
    auto last = std::unique(v.begin(), v.end());
    // v now holds {1 2 1 3 4 5 4 x x x}, where 'x' is indeterminate
    v.erase(last, v.end());
    print(2);

    // sort followed by unique, to remove all duplicates
    std::sort(v.begin(), v.end()); // {1 1 2 3 4 4 5}
    print(3);

    last = std::unique(v.begin(), v.end());
    // v now holds {1 2 3 4 5 x x}, where 'x' is indeterminate
    v.erase(last, v.end());
    print(4);
}

7.accumulate

vector<int> a = {0, 2, 3};
int sum = accumulate(a.begin(), a.end(), 0LL);
cout << sum << '\n';
//输出：5