Determine whether an integer is a palindrome. Do this without extra space.
if use recursive, like check the first dig and last dig, then remove them, check the rest, it will fail when digint like "1021", when remove the first and last one, the remaining would simply be "2", not "02".
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class
Solution {public: bool
isPalindrome(int
x) { //Palindrome means 回文 if(x < 0) return
false; if(x <= 9) return
true; int
t = x; int
n=0; while(t > 0) { n++; int
d = t%10; t=(t-d)/10; } int* s = new
int[n]; t = x; n=0; while(t > 0) { int
d = t%10; s[n++] = d; t=(t-d)/10; } for(int
i = 0; i<n/2;i++) { if(s[i] != s[n-i-1]) { return
false; } } return
true; }}; |
[LeetCode] [Palindrome Number 2012-01-04],布布扣,bubuko.com
[LeetCode] [Palindrome Number 2012-01-04]
原文：http://www.cnblogs.com/xxiao1119/p/3742662.html