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
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My own solve:
class Solution(object): def nextPermutation(self, nums): """ :type nums: List[int] :rtype: void Do not return anything, modify nums in-place instead. """ l0 = len(nums) if l0==2: if nums[1]>nums[0]: temp = nums[0] nums[0]=nums[1] nums[1]=temp elif l0>2: l1 = l0-3 flag = 0 while l1>=0: k=0 while (l0-k-l1)>=2: (arr,i) = self.subSort(nums[l1:l0-k]) if arr != []: nums[l1:l0-k] = arr ss = nums[l1+i+1:] ss.sort() nums[l1+i+1:]=ss flag = 1 break k += 1 if flag == 1: break l1 -= 1 if flag == 0: nums.sort() def subSort(self, arr): l = len(arr) while l>=2: l1 = l while l1>= 1: if arr[l-1] > arr[l1-1]: temp = arr[l1-1] arr[l1-1] = arr[l-1] arr[l-1] = temp return(arr,l1-1) l1 -= 1 l -= 1 return ([],0)
Others(http://bookshadow.com/weblog/2016/09/09/leetcode-next-permutation/):
解题思路：
首先从右向左遍历数组，找到第一个相邻的左<右的数对，记右下标为x，则左下标为x - 1
若x > 0，则再次从右向左遍历数组，直到找到第一个大于nums[x - 1]的数字为止，记其下标为y，交换nums[x - 1], nums[y]
最后将nums[x]及其右边的元素就地逆置
class Solution(object): def nextPermutation(self, nums): """ :type nums: List[int] :rtype: void Do not return anything, modify nums in-place instead. """ size = len(nums) for x in range(size - 1, -1, -1): if nums[x - 1] < nums[x]: break if x > 0: for y in range(size - 1, -1, -1): if nums[y] > nums[x - 1]: nums[x - 1], nums[y] = nums[y], nums[x - 1] break for z in range((size - x) / 2): nums[x + z], nums[size - z - 1] = nums[size - z - 1], nums[x + z]