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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.
将整数反转，然后与原来的数比较，如果相等则为Palindrome Number否则不是
/*********************************
* 日期：2015-01-20
* 作者：SJF0115
* 题目: 9.Palindrome Number
* 网址：https://oj.leetcode.com/problems/palindrome-number/
* 结果：AC
* 来源：LeetCode
* 博客：
**********************************/
#include <iostream>
using namespace std;
class Solution {
public:
bool isPalindrome(int x) {
// negative integer
if(x < 0){
return false;
}//if
int tmp = x;
int n = 0;
// reverse an integer
while(tmp){
n = n * 10 + tmp % 10;
tmp /= 10;
}//while
return (x == n);
}
};
int main(){
Solution solution;
int num = 1234321;
bool result = solution.isPalindrome(num);
// 输出
cout<<result<<endl;
return 0;
}
上一种思路，将整数反转时有可能会溢出。
这里的思路是不断的取第一位和最后一位（10进制）进行比较，相等则取第二位和倒数第二位.......直到完成比较或者中途不相等退出。
class Solution {
public:
bool isPalindrome(int x) {
// negative integer
if(x < 0){
return false;
}//if
// 位数
int divisor = 1;
while(x / divisor >= 10){
divisor *= 10;
}//while
int first,last;
while(x){
first = x / divisor;
last = x % 10;
// 高位和低位比较 是否相等
if(first != last){
return false;
}//if
// 去掉一个最高位和一个最低位
x = x % divisor / 10;
divisor /= 100;
}//while
return true;
}
};
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原文地址：http://blog.csdn.net/sunnyyoona/article/details/42926929
