leetcode 9 Palindrome Number 回文数

本文提供了一种不使用额外空间判断整数是否为回文数的方法,并提供了C++及Python的实现方案。讨论了负数及溢出情况的处理方式。

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

click to show spoilers.

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.

非常简洁的c++解决方案:
对于回文数只比较一半

public boolean isPalindrome1(int x) {
    if (x == 0) return true;
    // in leetcode, negative numbers and numbers with ending zeros
    // are not palindrome
    if (x < 0 || x % 10 == 0)
        return false;

    // reverse half of the number
    // the exit condition is y >= x
    // so that overflow is avoided.
    int y = 0;
    while (y < x) {
        y = y * 10 + (x % 10);
        if (x == y)  // to check numbers with odd digits
            return true;
        x /= 10;
    }
    return x == y; // to check numbers with even digits
}

python这个解法应该是前后分别对比:


class Solution:
    # @param x, an integer
    # @return a boolean
    def isPalindrome(self, x):
        if x < 0:
            return False

        ranger = 1
        while x / ranger >= 10:
            ranger *= 10

        while x:
            left = x / ranger
            right = x % 10
            if left != right:
                return False

            x = (x % ranger) / 10
            ranger /= 100

        return True

python字符串的解法:

class Solution:
    # @param {integer} x
    # @return {boolean}
    def isPalindrome(self, x):
        return str(x)==str(x)[::-1]
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