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Determine whether an integer is a palindrome. Do this without extra space.
Analyse：判断一个整数是否是回文数。注意负数都不是回文数。
1 class Solution { 2 public: 3 bool isPalindrome(int x) { 4 if(x < 0) return false; 5 string s; 6 while(x){ 7 char temp = x % 10 + ‘0‘; 8 s += temp; 9 x /= 10; 10 } 11 bool result = true; 12 for(int i = 0; i < s.length() / 2; i++){ 13 if(s[i] != s[s.length() - 1 - i]) return false; 14 } 15 return true; 16 } 17 };
或者：
1 class Solution { 2 public: 3 bool isPalindrome(int x) { 4 if(x < 0) return false; 5 else if(x < 10) return true; 6 else{ 7 int div = 1000000000; 8 while(x / div == 0) div /= 10; 9 bool result = true; 10 while(x){ 11 int left = x / div; 12 int right = x % 10; 13 if(left != right){ 14 result = false; 15 break; 16 } 17 x = (x- left * div) / 10; 18 div /= 100; 19 } 20 return result; 21 } 22 } 23 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4419442.html
