【LeetCode每天一题】Plus One(加一)

Given a non-empty array of digits representing a non-negative integer, plus one to the integer.The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.

Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.

思路

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  这道题在python中还是挺好解决的， 直接从列表的尾部开始进行并设置一个溢出标志量，然后执行加法操作，如果加之后的结果小于10的话，直接中断，并判断溢出标志量。否则指针减1，开始新一轮的计算。直到遍历到头部结束。时间复杂度为O(n), 空间复杂度为O(1)。
解决代码

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 class Solution(object):
     def plusOne(self, nums):
         """
         :type digits: List[int]
         :rtype: List[int]
         """
         index = len(nums)-1
         if not nums:
             return []
         flow = 0          # 溢出标志量
         while index >= 0:    # 从尾部开始向前遍历
             tem = nums[index] + 1
             if tem >= 10:     # 如果结果大于等于10的话，设置溢出标志量
                 flow = 1
                 nums[index] = tem %10
             else:                # 否则直接中断
                 nums[index] = tem %10
                 flow = 0
                 break
             index -= 1
         if flow == 1:          # 最后判断溢出是否为1， 因此可能会遇到比如999这种输入。
             nums.insert(0, 1)     # 在头部插入
         return nums