Palindrome Number
Palindrome Number

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

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Some hints:

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.

【注】问题要求O(1)空间复杂度，负数不是回文数

①每次提取头部和尾部的两个数，判断它们是否相等，判断后去掉头尾继续。采用除数和取余实现。暂时不能查看提交信息！耗时337ms。

public class Solution {
public boolean isPalindrome(int x) {
if(x < 0) return false;
int len = 1;
while(x / len >=10) len *= 10;
while(x > 0){
int head = x / len;//获取头部
int tail = x % 10;//获取尾部

return false;
}else{//去掉头部和尾部
x = (x % len) / 10;
len /= 100;

}
}
return true;
}
}

public class Solution { //206ms
public boolean isPalindrome(int x) {
if(x < 0) return false;
int div = 1;
while(x / div >= 10) div = div * 10;  //位数
while(x > 0) {
if(x / div != x % 10) return false;
x = x % div;  //去除头部
x = x / 10;  //去除尾部
div = div / 100;  //位数减2位
}
return true;
}
}

③ 用tempx保存数字x的头部，用reverse保存除头部之外x位数的翻转，对比整个x的头尾两位，以及中间所有位reverse以后是否与不reverse时一致，即可得到结果。

public class Solution { //207ms
public boolean isPalindrome(int x) {
if (x < 0) {
return false;
}
int tempX = x;
int reverse = 0;
while (tempX >= 10) { //// 当tempX剩下一位数的时候停下来
reverse = reverse * 10 + tempX % 10; //将除了最高位之外的位数翻转
tempX /= 10; //取得最高位
}

return tempX == x % 10 && reverse == x / 10; ////对比整个x的头尾两位，以及中间所有位reverse以后是否与不reverse时一致
}
}

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