No.9 Palindrome Number
Determine whether an integer is a palindrome. Do this without extra space.
Could negative integers be palindromes? (ie, -1)
If you are thinking of converting the integer to string, note the restriction of using extra space.
You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
There is a more generic way of solving this problem.
1 class Solution 2 { 3 public: 4 bool isPalindrome(int x) 5 {//判断整数是否为回文 6 //限制：不允许使用额外空间 7 //提示:负数是否为回文？；不能转换为string，因为不允许使用额外空间；反转数字的话，有可能溢出 8 //用纯数学的方法，想清楚就好，别慌，敢下手!!! 9 //参考：http://blog.csdn.net/linhuanmars/article/details/21145231 10 if(x < 0) 11 return false;//负数不是回文 12 if(x == 0) 13 return true; 14 15 int base = 1;//用于取高位 16 while(base <= x/10) 17 base *= 10; 18 19 while(x>0) 20 { 21 if(x/base != x%10) 22 return false; 23 24 x = (x%base)/10;//去首去尾 25 base /= 100;//每次消去两位 26 } 27 return true; 28 } 29 };
1 class Solution 2 { 3 public: 4 bool isPalindrome(int x) 5 {//判断整数是否为回文 6 //限制：不允许使用额外空间 7 //提示:负数是否为回文？；不能转换为string，因为不允许使用额外空间；反转数字的话，有可能溢出 8 //参考：http://www.cnblogs.com/remlostime/archive/2012/11/13/2767676.html 9 if(x < 0) 10 return false;//负数不是回文 11 if(x == 0) 12 return true; 13 14 int base = 1;//用于取高位 15 while(base <= x/10) 16 base *= 10; 17 18 int leftDigit ; 19 int rightDigit ; 20 while(x) 21 { 22 leftDigit = x/base; 23 rightDigit = x%10; 24 if(leftDigit != rightDigit) 25 return false; 26 27 x -= base*leftDigit;//去首 28 base /= 100;//每次消去两位 29 x /= 10;//去尾 30 } 31 return true; 32 } 33 };
原文：http://www.cnblogs.com/dreamrun/p/4569288.html
