std::is_permutation
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在头文件 <algorithm> 中定义
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template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(1) | (C++11 起) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first, ForwardIt1 last, |
(2) | (C++11 起) |
返回true的范围
[first1, last1)
,使该范围相等的范围内开始,d_first
中的元素,如果存在一个置换。第一个版本使用operator==
平等,第二个版本使用的二元谓词p
原文:
Returns true if there exists a permutation of the elements in the range
[first1, last1)
that makes that range equal to the range beginning at d_first
. The first version uses operator==
for equality, the second version uses the binary predicate p
目录 |
[编辑] 参数
first, last | - | 范围内的元素进行比较
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d_first | - | 用来比较的第二范围的开头
原文: the beginning of 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 Type1 &a, const Type2 &b); The signature does not need to have const &, but the function must not modify the objects passed to it. |
类型要求 | ||
-ForwardIt1, ForwardIt2 必须满足 ForwardIterator 的要求。
|
[编辑] 返回值
true如果范围
[first, last)
开始的范围内的一个排列在d_first
.原文:
true if the range
[first, last)
is a permutation of the range beginning at d_first
.[编辑] 复杂度
在最O(N2)应用的谓词,如果序列已经等于或完全N,其中N=std::distance(first, last).
原文:
At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N=std::distance(first, last).
[编辑] 可能的实现
template<class ForwardIt1, class ForwardIt2> bool is_permutation(ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first) { // skip common prefix std::tie(first, d_first) = std::mismatch(first, last, d_first); // iterate over the rest, counting how many times each element // from [first, last) appears in [d_first, d_last) if (first != last) { ForwardIt2 d_last = d_first; std::advance(d_last, std::distance(first, last)); for (ForwardIt1 i = first; i != last; ++i) { if (i != std::find(first, i, *i)) continue; // already counted this *i auto m = std::count(d_first, d_last, *i); if (m==0 || std::count(i, last, *i) != m) { return false; } } } return true; } |
[编辑] 示例
#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'; }
输出:
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
[编辑] 另请参阅
按字典顺序产生区间内元素下一个较大的排列组合 (函数模板) | |
按字典顺序产生区间内元素下一个较小的排列组合 (函数模板) |