LeetCode Java First 400 题解-009

Felomeng

于 2019-10-26 13:00:20 发布

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分类专栏：刷题文章标签： LeetCode

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本文介绍了一种高效判断整数是否为回文数的方法，无需额外空间，通过逆位比较实现，特别处理了负数及溢出情况。

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Palindrome Number Easy

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

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.

public boolean isPalindrome(int x) {

    if (x < 0 || (x != 0 && x % 10 == 0)) return false;

    int rev = 0;

    while (x > rev) {

        rev = rev * 10 + x % 10;

        x = x / 10;

    }

    return (x == rev || x == rev / 10);

}

思路：先判定特殊情况：负和个位数。如果够长，则把后面一半逆位反过来（止于逆反数大于前面一半时）逆反数（或逆反数/10）和前面一半对比即可。