[Leetcode] (python): 031-next permutation

Source of the topic:

https://leetcode.com/problems/next-permutation/

Test Instructions Analysis:

Enter an array. Output the next larger array of these number combinations than the input. If the input is the largest, then the output is the smallest array. For example, the 1,3,2 output. and 3,2,1 output.

Topic Ideas:

If there is a larger array than the input array nums, then there is bound to be nums[i] < Nums[j] (I < j). The smallest array larger than the Nums, then the first nums[i] < Nums[i + 1] When the last number is compared, the last number of nums[i] and I + 1 later than Nums[i], and then the "I" after the number of orders, then get the answer.

Code (Python):

　　

1 classSolution (Object):2 def nextpermutation (self, nums):3         """4: Type nums:list[int]5: Rtype:voidDo notreturnAnything, modify Numsinch-Place instead.6         """7Size =Len (nums)8         ifSize = =0or size = =1:9             returnTeni = size-2 One          whileI >=0: A             ifNums[i] < Nums[i +1]: -j = i +1 -                  whileJ <Size: the                     ifNums[i] >=Nums[j]: -                          Break -J + =1 -J-=1 +NUMS[I],NUMS[J] =Nums[j],nums[i] -Nums[i +1:] = sorted (nums[i +1:]) +                 return AI-=1 atMiddle = (Size-1)//2;k = 0 -          whileK <=Middle: -Nums[k],nums[size-1-K] = nums[size-1-K],nums[k] -K + =1 -         return

View Code

Reprint Please specify source: http://www.cnblogs.com/chruny/p/4918297.html

[Leetcode] (python): 031-next permutation