66. Plus One - 技术栈

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 = [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].

Constraints:

1 <= digits.length <= 100
0 <= digits[i] <= 9
digits does not contain any leading 0's.

链接: [66. Plus One

](https://leetcode.cn/problems/plus-one/description/

思路一：

这道题可以先求数组中的几个数构成的数字（使用[::-1]，加一，之后再成为数组。

python 复制代码

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        #return [int(i) for i in str(int(''.join([str(j) for j in digits]))+1)]
        a=[i*10**index for index,i in enumerate(digits[::-1])]
        num=sum(a)+1
        return [int(x) for  x in str(num)]

a='python'

b=a[::-1]

print(b) #nohtyp

c=a[::-2]

print© #nhy

#从后往前数的话，最后一个位置为-1

d=a[:-1] #从位置0到位置-1之前的数

print(d) #pytho

e=a[:-2] #从位置0到位置-2之前的数

print(e) #pyth

b = a[i:j] # 表示复制a[i]到a[j-1]，以生成新的list对象

a = [0,1,2,3,4,5,6,7,8,9]

b = a[1:3] # [1,2]

#当i缺省时，默认为0，即 a[:3]相当于 a[0:3]

#当j缺省时，默认为len(alist), 即a[1:]相当于a[1:10]

#当i,j都缺省时，a[:]就相当于完整复制一份a

b = a[i:j:s] # 表示：i,j与上面的一样，但s表示步进，缺省为1.

#所以a[i:j:1]相当于a[i:j]

#当s<0时，i缺省时，默认为-1. j缺省时，默认为-len(a)-1

#所以a[::-1]相当于 a[-1:-len(a)-1:-1]，也就是从最后一个元素到第一个元素复制一遍，即倒序。

思路二：

这道题可以先求数组中的几个数构成的数字（使用str，加一，之后再成为数组。

python 复制代码

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        return [int(i) for i in str(int(''.join([str(j) for j in digits]))+1)]

思路三:for循环遍历

python 复制代码

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        num = 0
        for n in range(len(digits)):
            num += (digits[len(digits)-n-1] * 10 ** n)
        return [int(n) for n in str(num+1)]