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)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 <stdio.h>
#include <stdlib.h>
//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", 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("\n");
for (int i=1; i<argc; i++) {
int n = atoi(argv[i]);
printf("%12d => %-12d %s!\n", n, reverse(n), reverse(reverse(n))==n ? "passed":"failed");
}
return 0;
}
原文：http://blog.csdn.net/cinmyheart/article/details/44788179
