STL algorithm演算法is_permutation(27)

標籤：algorithm   stl   演算法   c++11   

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);

該函數是用來判斷兩個序列是否為同一元素集的不同排列！

該函數使用operator==或者是pred來判斷兩個元素是否是相等的。

其行為類似：

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下面的codeblock居然不支援這個函數，我汗！

一個簡單的測試例子：

#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; elsecout<<"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; elsecout<<"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; elsecout<<"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; elsecout<<"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; elsecout<<"No ,v1 and ai is not permutation!"<<endl;     }

啟動並執行結果：

可以看到，雖然v3多了一個元素1,但是v1和v3還是屬於同一元素集的不同排列！

v4的資料類型為double，也是一樣！

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轉載請註明出處：http://blog.csdn.net/qq844352155

author:天下無雙

Email:[email protected]

2014-9-17

於GDUT

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STL algorithm演算法is_permutation(27)