LeetCode #Reverse Number#

刚背了单词，然后做个题玩玩～挑个软柿子踩踩～哈哈

很简单的思路．不过好玩的是我忘记检查处理完的数据是否符合整形数据返回了．因而好一会儿不能AC．

感谢 @Fantasy.　很快的指出我没有检查返回数据的范围．

先给出我超丑陋的解(python),　而后给出其他高手给出的很优雅的解！！也是用python

最后会给出利用java和C/C++的解．

Programmer : EOF
Date : 2015.03.31
File   :  reverse_interger.py


class Solution:
    # @return an integer
    def reverse(self, x):
		buffer =[]
		
		INT_MAX = 0x7fffffff

		INT_MIN = (-INT_MAX - 1)
		
		if x > INT_MAX or x < INT_MIN :
		    return 0

		if x > 0 :
			tmp = x
			counter = 1
			while tmp > 0 :
				counter += 1
				buffer.append(tmp % 10);
				tmp /= 10

			tmp = 0
			for i in range(0, len(buffer)) :
				tmp *= 10
				tmp += buffer[i]

			if tmp > INT_MAX or tmp < INT_MIN :
				return 0
		    
			return tmp

		elif x < 0 :
			tmp = -x
			counter = 1
			while tmp > 0 :
				counter += 1
				buffer.append(tmp % 10);
				tmp /= 10

			tmp = 0
			for i in range(0, len(buffer)) :
				tmp *= 10
				tmp += buffer[i]

			if tmp > INT_MAX or tmp < INT_MIN :
				return 0
		        
			return -tmp

		else:
			return 0

#--------------------------just for testing ----------------------------

s = Solution()

print s.reverse(123)

...　确实好搓就很简单做一下很基础的数学运算，利用余数和商的特性，把结果保存在list中．而后逆序的组合成整数，最后别忘记重新检查整数是否符合要求，是否越界！！

下面给出别人很优雅的解：

class Solution:
  # @return an integer
  def reverse(self, x):
    if x<0:
      sign = -1
    else:
      sign = 1
    strx=str(abs(x))
    r = strx[::-1]
    return sign*int(r)

简直不能再短．．帅的飞起

先强转成字符串，而后在利用python切片的技巧，逆序输出字符，然后再int()强转，限定好整数范围．

最后恢复数据的符号．

java解（来源于@凯旋冲锋）：利用了内置整形的特性

package reverse_integer;

public class Solution {
	public int reverse(int x) {
		long r = 0;
		while (x != 0) {
			r = r * 10 + x % 10;
			x /= 10;
		}
		return r > Integer.MAX_VALUE || r < Integer.MIN_VALUE ? 0 : (int) r;
	}

	public static void main(String[] args) {
		System.out.println(new Solution().reverse(1563847412));
	}
}

最后献上皓神的解答，C语言实现．

自己抽自己一巴掌，擦，居然没看清楚皓神写的注释！！

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer,
?

// Source : https://oj.leetcode.com/problems/reverse-integer/
// Author : Hao Chen
// Date   : 2014-06-18

/********************************************************************************** 
* 
* Reverse digits of an integer.
* 
* Example1: x =  123, return  321
* Example2: x = -123, return -321
* 
* 
* Have you thought about this?
* 
* Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
* 
* > If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
* 
* > Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, 
*   then the reverse of 1000000003 overflows. How should you handle such cases?
* 
* > Throw an exception? Good, but what if throwing an exception is not an option? 
*   You would then have to re-design the function (ie, add an extra parameter).
* 
*               
**********************************************************************************/

#include 
  
   
#include 
   
     //Why need the INT_MIN be defined like that? //Please take a look: // http://stackoverflow.com/questions/14695118/2147483648-0-returns-true-in-c #define INT_MAX 2147483647 #define INT_MIN (-INT_MAX - 1) int reverse(int x) { int y=0; int n; while( x != 0){ n = x%10; //Checking the over/underflow. //Actually, it should be y>(INT_MAX-n)/10, but n/10 is 0, so omit it. if (y > INT_MAX/10 || y < INT_MIN/10){ return 0; } y = y*10 + n; x /= 10; } return y; } #define TEST(n, e) printf(%12d => %-12d %s! , n, reverse(n), e == reverse(n)?passed:failed) int main(int argc, char**argv) { //basic cases TEST( 123, 321); TEST( -123, -321); TEST( -100, -1); TEST( 1002, 2001); //big integer TEST( 1463847412, 2147483641); TEST(-2147447412, -2147447412); TEST( 2147447412, 2147447412); //overflow TEST( 1000000003, 0); TEST( 2147483647, 0); TEST(-2147483648, 0); //customized cases if (argc<2){ return 0; } printf( ); for (int i=1; i
    
      %-12d %s! , n, reverse(n), reverse(reverse(n))==n ? passed:failed); } return 0; }

LeetCode #Reverse Number#(一)