Determine whether an integer is a palindrome. Do this without extra space.

尝试用两头分别比较的方法,结果发现无法解决1000021这种问题

 public class Solution {
public boolean isPalindrome(int x) {
if(x<0) return false;
int lastbit=x%10;
int firstbit;
int num=x;
int bits=0;
while(num/10!=0){
num=num/10;
bits++;
}
firstbit=num;
if(firstbit!=lastbit) return false;
else{
x=x-firstbit*((int)(Math.pow(10,bits)));
x=x/10;
if(x==0) return true;
else return isPalindrome(x);
}
}
}

查看了论坛的解答,看到一个很好的solution, 它怎么想到13行的 div/=100的,给跪了,这样刚好解决了中间有0的问题,比如1021会被判定为false, 而121会被判定为true。 解答详见:http://leetcode.com/2012/01/palindrome-number.html

 public class Solution {
public boolean isPalindrome(int x) {
if (x < 0) return false;
int div = 1;
while (x / div >= 10) {
div *= 10;
}
while (x != 0) {
int l = x / div;
int r = x % 10;
if (l != r) return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
}

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