[LeetCode] 9. Palindrome Number

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Could you solve it without converting the integer to a string?

回文数。

题意是判断一个数字是否是回文数。这个题跟第7题非常类似，有两种做法，一种跟第七题很类似，另一种是把 input 数字转换成字符串然后双指针逼近检测是否有不一样的字符。

无论什么做法，都有一些需要去掉的 corner case，首先负数一定不是回文数，能被 10 整除的数字也不是，其次回文数一定不会越界。类似第七题的代码如下，

时间O(n)

空间O(1)

JavaScript实现

/**
 * @param {number} x
 * @return {boolean}
 */
var isPalindrome = function (x) {
    // corner case
    if (x < 0 || (x !== 0 && x % 10 === 0)) return false;

    // normal case
    let palindrome = x;
    let reverse = 0;
    while (x > 0) {
        reverse = reverse * 10 + (x % 10);
        x = ~~(x / 10);
    }
    return palindrome === reverse;
};

Java实现

class Solution {
    public boolean isPalindrome(int x) {
        // corner case
        if (x < 0 || (x != 0 && x % 10 == 0)) return false;

        // normal case
        int palindrome = x;
        int reverse = 0;
        while (x > 0) {
            reverse = reverse * 10 + x % 10;
            x /= 10;
        }
        return palindrome == reverse;
    }
}

two pointer 逼近，如果发现两边不一样即可return false；否则就return true。

时间O(logn)

空间O(1)

Java实现

class Solution {
    public boolean isPalindrome(int x) {
        String s = String.valueOf(x);
        int left = 0;
        int right = s.length() - 1;
        // corner case
        if (s.charAt(0) == '-') {
            return false;
        }
        
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            } else {
                left++;
                right--;
            }
        }
        return true;
    }
}

LeetCode 题目总结