std::is_permutation
Defined in header <algorithm> | ? | ? |
|---|---|---|
template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 ); | (1) | (since C++11) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p ); | (2) | (since C++11) |
template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2 ); | (3) | (since C++14) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2, BinaryPredicate p ); | (4) | (since C++14) |
回报true如果存在范围内的元素排列[first1, last1)，这使得该范围等于范围。[first2,last2)，在哪里last2表示first2 + (last1 - first1)如果没有人给的话。
1，3%29元素的比较operator==如果行为不是等价关系...
使用给定的二进制谓词对2，4%29元素进行比较p如果行为不是等价关系，则行为是未定义的。
参数
first1, last1 | - | the range of elements to compare |
|---|---|---|
first2, last2 | - | the second range to compare |
p | - | binary predicate which returns ?true if the elements should be treated as equal. The signature of the predicate function should be equivalent to the following: bool pred(const Type &a, const Type &b); Type should be the value type of both ForwardIt1 and ForwardIt2. The signature does not need to have const &, but the function must not modify the objects passed to it. ? |
类型要求
-前进1，前进2必须符合先行者的要求。
-ForwardIt 1，ForwardIt 2必须具有相同的值类型。
返回值
true如果范围[first1, last1)是范围的排列[first2, last2)...
复杂性
最多为O%28N2%29的谓词应用程序，或者如果序列已经相等，则精确N，其中N=std::distance(first1, last1)...
但是如果ForwardIt1和ForwardIt2满足…的要求RandomAccessIterator和std::distance(first1, last1)!=std::distance(first2, last2)不进行谓词的应用。
可能的实施
模板<类ForwardIt 1，类ForwardIt 2>bool是[医]排列%28ForwardIt1优先，ForwardIt 1最后，ForwardIt2d[医]第一%29{//跳过通用前缀STD：：Tie%281，d[医]第一%29=STD：：不匹配%281，最后，d[医]第一个%29；//迭代其余的元素，计算每个元素//从元素的次数。[首先，最后%29出现在。[丁[医]第一，d[医]最后%29如果%28First%21=最后%29(ForwardIt2d)[医]最后=d[医]第一；性病：：预付%28d[医]最后，std：：距离%281，最后%29%29；对于%28 ForwardIt 1 i=先；i%21=Lest；++i%29{if%28i%21=std：：查找%28First，i，%2AI%29%29继续；//已计算此%2Ai AUTO m=STD：：Count%28d[医]第一，d[医]最后，%2Ai%29；if%28m=0最后，数%28 i，%2AI%29%21=m%29{返回false；}}返回true；}
*。
例
二次
#include <algorithm>
#include <vector>
#include <iostream>
int main()
{
std::vector<int> v1{1,2,3,4,5};
std::vector<int> v2{3,5,4,1,2};
std::cout << "3,5,4,1,2 is a permutation of 1,2,3,4,5? "
<< std::boolalpha
<< std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n';
std::vector<int> v3{3,5,4,1,1};
std::cout << "3,5,4,1,1 is a permutation of 1,2,3,4,5? "
<< std::boolalpha
<< std::is_permutation(v1.begin(), v1.end(), v3.begin()) << '\n';
}二次
产出：
二次
3,5,4,1,2 is a permutation of 1,2,3,4,5? true
3,5,4,1,1 is a permutation of 1,2,3,4,5? false二次
另见
next_permutation | generates the next greater lexicographic permutation of a range of elements (function template) |
|---|---|
prev_permutation | generates the next smaller lexicographic permutation of a range of elements (function template) |
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