std::prev_permutation
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| Defined in header <algorithm>
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| template< class BidirIt > bool prev_permutation( BidirIt first, BidirIt last); |
(1) | |
| template< class BidirIt, class Compare > bool prev_permutation( BidirIt first, BidirIt last, Compare comp); |
(2) | |
变换的范围内
[first, last)到先前从该组的所有字典顺序排列的排列相对于operator<或comp置换。返回true如果这样的排列存在，否则将到最后置换（如果由std::sort(first, last); std::reverse(first, last);），并返回false.Original:
Transforms the range
[first, last) into the previous permutation from the set of all permutations that are lexicographically ordered with respect to operator< or comp. Returns true if such permutation exists, otherwise transforms the range into the last permutation (as if by std::sort(first, last); std::reverse(first, last);) and returns false.The text has been machine-translated via Google Translate.
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目录 |
[编辑] 参数
| first, last | - | 元素的范围内的重排
Original: the range of elements to permute The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
| comp | - | comparison function which returns true if the first argument is less than the second. The signature of the comparison function should be equivalent to the following:
The signature does not need to have const &, but the function must not modify the objects passed to it. | |||||||||
| Type requirements | |||||||||||
-BidirIt must meet the requirements of ValueSwappable and BidirectionalIterator.
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[编辑] 返回值
true如果之前的旧字典顺序排列。 false如果第一置换达到的范围内，被复位到最后置换.
Original:
true if the new permutation precedes the old in lexicographical order. false if the first permutation was reached and the range was reset to the last permutation.
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[编辑] 复杂性
在大多数
(last-first)/2掉期.Original:
At most
(last-first)/2 swaps.The text has been machine-translated via Google Translate.
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[编辑] 可能的实现
template<class BidirIt> bool prev_permutation(BidirIt first, BidirIt last) { if (first == last) return false; BidirIt i = last; if (first == --i) return false; while (1) { BidirIt i1, i2; i1 = i; if (*i1 < *--i) { i2 = last; while (!(*--i2 < *i)) ; std::iter_swap(i, i2); std::reverse(i1, last); return true; } if (i == first) { std::reverse(first, last); return false; } } } |
[编辑] 为例
下面的代码打印出所有的字符串“ABC”以相反的顺序排列
Original:
The following code prints all six permutations of the string "abc" in reverse order
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#include <algorithm> #include <string> #include <iostream> #include <functional> int main() { std::string s="abc"; std::sort(s.begin(), s.end(), std::greater<char>()); do { std::cout << s << ' '; } while(std::prev_permutation(s.begin(), s.end())); std::cout << '\n'; }
Output:
cba cab bca bac acb abc
[编辑] 另请参阅
| （C++11） |
determines if a sequence is a permutation of another sequence （函数模板） |
| prev_permutation |
generates the next smaller lexicographic permutation of a range of elements （函数模板） |
