STL algorithm算法lexicographical_compare(30)
lexicographical_compare原型:
std::lexicographical_compare
| default (1) |
template <class InputIterator1, class InputIterator2> |
|---|---|
| custom (2) |
template <class InputIterator1, class InputIterator2, class Compare> |
该函数是依照字典序測试[frist1,last1)是否小于[first2,last2).
字典序是指依照字母在字典中出现的顺序。
该函数使用opeartor<或者是comp进行比較。
假设两个序列长度不同,而且短序列和长序列头部全然一样,比如example和examplee.那么,长度大的字典序比短序的大。
其行为类似于:
|
|
一个简单的样例:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(int argv,char **argc)
{
vector<char> v1{'h','e','l','l','o'};
vector<char> v2{'h','e','l','l','o','o'};
vector<char> v3{'h','e','l','m','o'};
cout<<"v1=";
for(char i:v1)
cout<<i<<" ";
cout<<endl;
cout<<"v2=";
for(char i:v2)
cout<<i<<" ";
cout<<endl;
cout<<"v3=";
for(char i:v3)
cout<<i<<" ";
cout<<endl; if(lexicographical_compare(v1.begin(),v1.end(),v2.begin(),v2.end()))
cout<<"v1 is less than v2 "<<endl;
else
cout<<"v2 is less than v1 "<<endl; if(lexicographical_compare(v1.begin(),v1.end(),v3.begin(),v3.end()))
cout<<"v1 is less than v3 "<<endl;
else
cout<<"v3 is less than v1 "<<endl; }
执行截图:
该函数是否仅仅能比較字母呢?答案是肯定的,不是!
由于对于随意的能够使用opeartor<进行比較的对象都能够使用该函数!
一个简单的样例:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(int argv,char **argc)
{
vector<int> v1{1,2,3,4};
vector<int> v2{1,2,3,4,5};
vector<int> v3{1,2,3,3};
cout<<"v1=";
for(int i:v1)
cout<<i<<" ";
cout<<endl;
cout<<"v2=";
for(int i:v2)
cout<<i<<" ";
cout<<endl;
cout<<"v3=";
for(int i:v3)
cout<<i<<" ";
cout<<endl; if(lexicographical_compare(v1.begin(),v1.end(),v2.begin(),v2.end()))
cout<<"v1 is less than v2 "<<endl;
else
cout<<"v2 is less than v1 "<<endl; if(lexicographical_compare(v1.begin(),v1.end(),v3.begin(),v3.end()))
cout<<"v1 is less than v3 "<<endl;
else
cout<<"v3 is less than v1 "<<endl; }
执行截图:
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//写的错误或者不好的地方请多多指导,能够在以下留言或者点击左上方邮件地址给我发邮件,指出我的错误以及不足,以便我改动,更好的分享给大家,谢谢。
转载请注明出处:http://blog.csdn.net/qq844352155
author:天下无双
Email:coderguang@gmail.com
2014-9-17
于GDUT
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