小知识，大挑战！本文正在参与“程序员必备小知识”创作活动。
描述
You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0's.
Increment the large integer by one and return the resulting array of digits.
Example 1:
Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].
Example 2:
Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].
Example 3:
Input: digits = [0]
Output: [1]
Explanation: The array represents the integer 0.
Incrementing by one gives 0 + 1 = 1.
Thus, the result should be [1].
Example 4:
Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].
Note:
1 <= digits.length <= 100
0 <= digits[i] <= 9
digits does not contain any leading 0's.
解析
根据题意，给出了一个列表表示一个整数，然后给这个“整数”加一，然后返回所得结果的列表形式。其实思路比较简单，就是模拟正常的整数相加规律，从后往前相加，如果满 10 就往前一位进一，需要注意的是如果在最左边的元素为 10 ，需要特殊处理。
解答
class Solution(object):
def plusOne(self, digits):
"""
:type digits: List[int]
:rtype: List[int]
"""
digits[-1] += 1
for i in range(len(digits)-1, 0, -1):
if digits[i] != 10:
break
digits[i] = 0
digits[i-1] += 1
if digits[0] == 10:
digits[0] = 0
return [1] + digits
return digits
运行结果
Runtime: 39 ms, faster than 6.82% of Python online submissions for Plus One.
Memory Usage: 13.2 MB, less than 92.25% of Python online submissions for Plus One.
解析
当然还可以使用内置函数，先将 digits 换算成整数然后加一，最后将结果用列表的形式展示出来即可。
解答
class Solution(object):
def plusOne(self, digits):
"""
:type digits: List[int]
:rtype: List[int]
"""
num = 0
N = len(digits)
for i, digit in enumerate(digits):
num += pow(10, N-i-1) * digit
return [int(i) for i in str(num+1)]
运行结果
Runtime: 36 ms, faster than 10.12% of Python online submissions for Plus One.
Memory Usage: 13.5 MB, less than 42.47% of Python online submissions for Plus One.
解析
看了官网的大神解答，才发现还能用递归进行求解，是我格局小了。
解答
class Solution(object):
def plusOne(self, digits):
"""
:type digits: List[int]
:rtype: List[int]
"""
if digits[-1] + 1 < 10:
digits[-1] += 1
return digits
elif len(digits) == 1 and digits[-1] + 1 == 10:
return [1, 0]
else:
digits[-1] = 0
digits[:-1] = self.plusOne(digits[:-1])
return digits
运行结果
Runtime: 32 ms, faster than 15.85% of Python online submissions for Plus One.
Memory Usage: 13.4 MB, less than 42.47% of Python online submissions for Plus One.
原题链接：leetcode.com/problems/pl…
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