31. Next Permutation
- Total Accepted: 87393
- Total Submissions: 313398
- Difficulty: Medium
- Contributors: Admin
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.1,2,3 → 1,3,23,2,1 → 1,2,31,1,5 → 1,5,1
Code
class Solution(object):
def nextPermutation(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
# find the last number that nums[i] > nums[i-1]
i = len(nums)-1
while i > 0:
if nums[i] > nums[i-1]:
break
i -= 1
# no nums[i] > nums[i-1] found. Reverse whole array
if i == 0:
self.reverse_sort(nums, 0, len(nums))
return
# find the last number that nums[j] > nums[i-1]
for j in xrange(len(nums)-1, i-1, -1):
if nums[j] > nums[i-1]:
nums[i-1], nums[j] = nums[j], nums[i-1]
break
self.reverse_sort(nums, i, len(nums))
def reverse_sort(self, nums, start, end):
"""
Reverse the order in the range
start: inclusive, end: exclusive
"""
mid = start + (end - start) / 2
for i in xrange(start, mid):
j = len(nums)-1-i+start
nums[i], nums[j] = nums[j], nums[i]
Idea
See comments and the plot for an example 125431:
Reference: https://discuss.leetcode.com/topic/2542/share-my-o-n-time-solution