Leetcode题解:Palindrome Number(回文数)

原题:Palindrome Number
难度:Easy

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:

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

对于判断一个数是否是回文数,一种简单的方法是将数字的各位存放在数组中,然后从数字的头和尾开始比较。

代码如下:

// golang
// leetcode测试成绩: 56ms  超过100%的解法
func isPalindrome(x int) bool {
    if x < 0 {
        return false
    } else if x == 0 {
        return true
    } 
    i := 0
    var buf [22]int
    // 将数字逐位存入数组
    for ;x > 0;i++ {
        buf[i] = x % 10
        x = x / 10
    }
    length := i
    // 逐位比较
    for i = 0;i < length / 2;i++ {
        if buf[i] != buf[length - 1 - i] {
            return false
        }
    }
    return true
}

上面这种方法比较简单,与回文字符串的判断基本一致。还有一种方法是按相相反的顺序构造出新数字(如123 -> 321),并与原数字比较。

// golang
// leetcode测试成绩: 56ms  超过100%的解法
func isPalindrome(x int) bool {
    if x >= 0 && x < 10{
        return true
    } else if x < 0 || x % 10 == 0{
        return false
    }
    new_num := 0
    for x > new_num  {
        new_num = new_num * 10 + x % 10 
        x = x / 10 
    }
    return (x == new_num || x == new_num / 10 )
}

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