C++primer 练习10.16

// 10.3.2.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;

string &  make_plural(int cnt, string &str)
{
     str=cnt > 1 ? (str + "s") : str;
     return str;
}

void biggies(vector<string> &words, vector<string>::size_type sz)
{
    sort(words.begin(), words.end());
    //按照默认形式对容器中的元素排序

    auto uniqIte = unique(words.begin(), words.end());
    //将重复的元素放到迭代器范围的后面，并且返回第一个重复元素的迭代器

    words.erase(uniqIte,words.end());
    //删除重复的元素
    //stable_sort(words.begin(), words.end(), [](const string &s1, const string &s2) {return s1.size() < s2.size();});
    //这个是书上的写法

    auto ite=partition(words.begin(), words.end(), [sz](const string &s) {return s.size() >= sz;});
    //返回第一个迭代器范围内不满足lambda表达式的元素，并且将满足表达式的元素放在范围前面，不满足的放在后面
    unsigned int cnt = 0;
    for (auto beg = words.begin();beg != ite;++beg, ++cnt);
    //得到长度不小于sz的字符串的个数
    
    cout << cnt << " " << make_plural(cnt, string("word")) << " of length " <<
        sz << " or longer" << endl;


    for_each(words.begin(), ite, [](const string &s) {cout << s << " ";});
    //输出迭代器范围内的元素（即长度不小于sz的元素）
    cout << endl;
}

int main()
{
    vector<string> strVec = { "lidanda","hahaha","liyou","wumei","ni","sansan" };
    biggies(strVec, 6);
    return 0;
}