LeetCode 7. Reverse Integer
https://leetcode.com/problems/reverse-integer/description/
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123 Output: 321
Example 2:
Input: -123 Output: -321
Example 3:
Input: 120 Output: 21
Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
- 简单题,类似字符串处理。主要是考虑integer overflow情况。比较tricky的方法是用long来存储result结果,然后判断是否超过INT_MAX/INT_MIN。
- 我用的还是判断overflow。由于输入不可能是INT_MAX/INT_MIN的reverse,所以不需要考虑abs(INT_MIN) > abs(INT_MAX)的情况。
1 // 2 // main.cpp 3 // LeetCode 4 // 5 // Created by Hao on 2017/3/16. 6 // Copyright © 2017年 Hao. All rights reserved. 7 // 8 9 #include <iostream> 10 #include <cstring> 11 #include <vector> 12 using namespace std; 13 14 class Solution { 15 public: 16 int reverse(int x) { 17 int result = 0; 18 int sign = 1; 19 20 if (x < 0) { 21 sign = -1; 22 x = abs(x); 23 } 24 25 while (x != 0) { 26 if (result > (INT_MAX - x % 10) / 10) // it's ok coz input would not be reverse of INT_MIN 27 return 0; 28 29 result = result * 10 + x % 10; 30 x /= 10; 31 } 32 33 return result * sign; 34 } 35 }; 36 37 int main(int argc, char* argv[]) 38 { 39 Solution testSolution; 40 string result; 41 42 vector<int> iVec = {123, -123, 120, 2147483647, -2147483648, 0}; 43 44 /* 45 321 46 -321 47 21 48 0 49 0 50 0 51 */ 52 for (auto i : iVec) { 53 cout << testSolution.reverse(i) << endl; 54 } 55 56 return 0; 57 }
- 一种解法同上,字符串处理。注意Python3中整除用//,而不是/。
- 第二种解法利用Python sliding s[::-1]做字符串反转。
- Python3 字符串 | 菜鸟教程
- http://www.runoob.com/python3/python3-string.html
- python - Maximum and Minimum values for ints - Stack Overflow
- https://stackoverflow.com/questions/7604966/maximum-and-minimum-values-for-ints
- Python 3
- In Python 3, this question doesn't apply. The plain
int
type is unbounded. - However, you might actually be looking for the machine's word size. That's still available in Python 3 as
sys.maxsize
.
- In Python 3, this question doesn't apply. The plain
- Python 2
- In Python 2, the maximum value for plain
int
values is available assys.maxint
: >>> sys.maxint 9223372036854775807
- In Python 2, the maximum value for plain
- You can calculate the minimum value with
-sys.maxint - 1
as shown here. - Python seamlessly switches from plain to long integers once you exceed this value. So most of the time, you won't need to know it.
1 class Solution: 2 def reverse(self, x: int) -> int: 3 result = 0 4 sign = 1 5 6 if x < 0: 7 sign = -1 8 x = abs(x) 9 10 while x != 0: 11 if result > ( 2**31 - ( x % 10 ) ) // 10: # be careful of floor division (integer division) 12 return 0 13 14 result = result * 10 + x % 10 15 x //= 10 # be careful of floor division (integer division) 16 17 return result * sign 18 19 def reverse2(self, x: int) -> int: 20 # [begin:end:step] - leaving begin and end off and specifying a step of -1, it reverses a string. 21 s = str( abs( x ) ) 22 x_reverse = int( s[ ::-1 ] ) if x > 0 else (- int( s[ ::-1 ] ) ) 23 24 if ( ( x_reverse >= - 2**31 ) and ( x_reverse <= 2**31 - 1 ) ): 25 return x_reverse 26 else: 27 return 0