题目 #

Given an integer x, return true if x is palindrome integer.

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

• For example, 121 is a palindrome while 123 is not.

思路1 #

• 最简单直接的想法，转成字符串，然后判断

 1func isPalindrome(x int) bool {
 2	s := fmt.Sprint(x)
 3	i := 0
 4	j := len(s) - 1
 5	for i < j {
 6		if s[i] != s[j] {
 7			return false
 8		}
 9		i++
10		j--
11	}
12	return true
13}

思路2 #

• 无情打脸，官方解法nb
• 转字符串太占内存，直接反转一半（位数为奇数多反转一个）的数字，然后比较

位数为奇数

@startuml
rectangle x {
    (121) as left1
    (12) as left2
    (1) as left3
}

rectangle revert {
    ( ) as right1
    (1) as right2
    (12) as right3
}

(x == revert/10) as mid

left1 --> left2
left2 --> left3

right1 --> right2
right2 --> right3

left3 --> mid
right3 --> mid
@enduml

位数为偶数

@startuml
rectangle x {
    (1221) as left1
    (122) as left2
    (12) as left3
}

rectangle revert {
    ( ) as right1
    (1) as right2
    (12) as right3
}

(x == revert) as mid

left1 --> left2
left2 --> left3

right1 --> right2
right2 --> right3

left3 --> mid
right3 --> mid
@enduml

代码实现

 1func isPalindrome(x int) bool {
 2	if x == 0 {
 3		return true
 4	}
 5	if x < 0 || x%10 == 0 {
 6		return false
 7	}
 8	revertNum := 0
 9	for x > revertNum {
10		revertNum = revertNum*10 + x%10
11		x /= 10
12	}
13	return x == revertNum || x == revertNum/10
14}