The most obvious idea is to maintain two divisors to get the most and least significant digits and compare them. Well, there are much more clever ideas, like this one, whose code is rewritten below. Play with it :-)
1 class Solution { 2 public: 3 bool isPalindrome(int x) { 4 if (x < 0 || (x && !(x % 10))) return false; 5 int s = 0; 6 while (x > s) { 7 s = s * 10 + x % 10; 8 x /= 10; 9 } 10 return x == s || x == s / 10; 11 } 12 };
原文：http://www.cnblogs.com/jcliBlogger/p/4732539.html