标签:dex i+1 src return lis ima self sel int
题目描述:
方法一:O(n) O(1)
class Solution: def nextPermutation(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ firstIndex = -1 n = len(nums) def reverse(nums,i,j): while i<j: nums[i],nums[j] = nums[j],nums[i] i += 1 j -= 1 return nums for i in range(n-2, -1, -1): if nums[i] < nums[i+1]: firstIndex = i break if firstIndex == -1: reverse(nums,0,n-1) return secondIndex = -1 for i in range(n-1, firstIndex, -1): if nums[i] > nums[firstIndex]: secondIndex = i break nums[firstIndex],nums[secondIndex] = nums[secondIndex], nums[firstIndex] reverse(nums, firstIndex+1, n-1)
标签:dex i+1 src return lis ima self sel int
原文地址:https://www.cnblogs.com/oldby/p/11164677.html