SGI

next_permutation

Category: algorithms Component type: function

Prototype

Next_permutation is an overloaded name; there are actually two next_permutation functions.
template <class BidirectionalIterator>
bool next_permutation(BidirectionalIterator first,
                      BidirectionalIterator last);

template <class BidirectionalIterator, class StrictWeakOrdering>
bool next_permutation(BidirectionalIterator first, BidirectionalIterator last,
                      StrictWeakOrdering comp);

Description

Next_permutation transforms the range of elements [first, last) into the lexicographically next greater permutation of the elements. There is a finite number of distinct permutations (at most N! [1], where N is last - first), so, if the permutations are ordered by lexicographical_compare, there is an unambiguous definition of which permutation is lexicographically next. If such a permutation exists, next_permutation transforms [first, last) into that permutation and returns true. Otherwise it transforms [first, last) into the lexicographically smallest permutation [2] and returns false.

The postcondition is that the new permutation of elements is lexicographically greater than the old (as determined by lexicographical_compare) if and only if the return value is true.

The two versions of next_permutation differ in how they define whether one element is less than another. The first version compares objects using operator<, and the second compares objects using a function object comp.

Definition

Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h.

Requirements on types

For the first version, the one that takes two arguments: For the second version, the one that takes three arguments:

Preconditions

Complexity

Linear. At most (last - first) / 2 swaps.

Example

This example uses next_permutation to implement the worst known deterministic sorting algorithm. Most sorting algorithms are O(N log(N)), and even bubble sort is only O(N^2). This algorithm is actually O(N!).
template <class BidirectionalIterator> 
void snail_sort(BidirectionalIterator first, BidirectionalIterator last)
{
  while (next_permutation(first, last)) {}
}

int main()
{
  int A[] = {8, 3, 6, 1, 2, 5, 7, 4};
  const int N = sizeof(A) / sizeof(int);

  snail_sort(A, A+N);
  copy(A, A+N, ostream_iterator<int>(cout, "\n"));
}

Notes

[1] If all of the elements in [first, last) are distinct from each other, then there are exactly N! permutations. If some elements are the same as each other, though, then there are fewer. There are, for example, only three (3!/2!) permutations of the elements 1 1 2.

[2] Note that the lexicographically smallest permutation is, by definition, sorted in nondecreasing order.

See also

prev_permutation, lexicographical_compare, LessThan Comparable, Strict Weak Ordering, sort
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