STL algorithm算法is_permutation(27)

is_permutation原型：

std::is_permutation

equality (1)	template <class ForwardIterator1, class ForwardIterator2> bool is_permutation (ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2);
predicate (2)	template <class ForwardIterator1, class ForwardIterator2, class BinaryPredicate> bool is_permutation (ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, BinaryPredicate pred);

该函数是用来推断两个序列是否为同一元素集的不同排列。

该函数使用operator==或者是pred来推断两个元素是否是相等的。

其行为类似：

template <class InputIterator1, class InputIterator2>
bool is_permutation (InputIterator1 first1, InputIterator1 last1,
                       InputIterator2 first2)
{
  std::tie (first1,first2) = std::mismatch (first1,last1,first2);
if (first1==last1) return true;
  InputIterator2 last2 = first2; std::advance (last2,std::distance(first1,last1));
for (InputIterator1 it1=first1; it1!=last1; ++it1) {
if (std::find(first1,it1,*it1)==it1) {
auto n = std::count (first2,last2,*it1);
if (n==0 || std::count (it1,last1,*it1)!=n) return false;
    }
  }
return true;
}

坑的是windows以下的codeblock竟然不支持这个函数，我汗！

一个简单的測试样例：

#include <iostream>
#include <vector>
#include <array>
#include <algorithm>
using namespace std;
int main(){
    vector<int> v1{1,2,3,4,5,6};
    vector<int> v2{1,5,6,4,3,2};
    vector<int> v3{1,3,2,5,6,4,1};//add a elements 1
    vector<double> v4{1,2,4,3,5,6};
	vector<int> v5{1,0,2,3,4,5,6};
    array<int,6> ai{6,5,3,4,2,1};  

    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;
   cout<<"v4=";
    for(double &i:v4)
        cout<<i<<" ";
    cout<<endl;
	 cout<<"v5=";
    for(int &i:v5)
        cout<<i<<" ";
    cout<<endl;
   cout<<"ai=";
    for(int &i:ai)
        cout<<i<<" ";
    cout<<endl;
    if(is_permutation(v1.begin(),v1.end(),v2.begin()))
		cout<<"Yes ,v1 and v2 is permutation!"<<endl;
	else
		cout<<"No ,v1 and v2 is not permutation!"<<endl; 		 

	if(is_permutation(v1.begin(),v1.end(),v3.begin()))
		cout<<"Yes ,v1 and v3 is permutation!"<<endl;
	else
		cout<<"No ,v1 and v3 is not permutation!"<<endl; 

	if(is_permutation(v1.begin(),v1.end(),v4.begin()))
		cout<<"Yes ,v1 and v4 is permutation!"<<endl;
	else
		cout<<"No ,v1 and v4 is not permutation!"<<endl; 

	if(is_permutation(v1.begin(),v1.end(),v5.begin()))
		cout<<"Yes ,v1 and v5 is permutation!"<<endl;
	else
		cout<<"No ,v1 and v5 is not permutation!"<<endl; 

	if(is_permutation(v1.begin(),v1.end(),ai.begin()))
		cout<<"Yes ,v1 and ai is permutation!"<<endl;
	else
		cout<<"No ,v1 and ai is not permutation!"<<endl; 

}

执行的结果：

能够看到，尽管v3多了一个元素1,可是v1和v3还是属于同一元素集的不同排列！

v4的数据类型为double，也是一样！

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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:[e-mail]

2014-9-17

于GDUT

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