LeetCode-in-All

31. Next Permutation

Medium

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).

The replacement must be in place and use only constant extra memory.

Example 1:

Input: nums = [1,2,3]

Output: [1,3,2]

Example 2:

Input: nums = [3,2,1]

Output: [1,2,3]

Example 3:

Input: nums = [1,1,5]

Output: [1,5,1]

Example 4:

Input: nums = [1]

Output: [1]

Constraints:

Solution

# @param {Integer[]} nums
# @return {Void} Do not return anything, modify nums in-place instead.
def next_permutation(nums)
  return if nums.nil? || nums.length <= 1

  i = nums.length - 2
  while i >= 0 && nums[i] >= nums[i + 1]
    i -= 1
  end

  if i >= 0
    j = nums.length - 1
    while nums[j] <= nums[i]
      j -= 1
    end
    swap_next(nums, i, j)
  end

  reverse_next(nums, i + 1, nums.length - 1)
end

private

def swap_next(nums, i, j)
  temp = nums[i]
  nums[i] = nums[j]
  nums[j] = temp
end

def reverse_next(nums, i, j)
  while i < j
    swap_next(nums, i, j)
    i += 1
    j -= 1
  end
end