https://leetcode.com/problems/palindrome-number/description/
palindrome-number： 回文數(shù) 技健， 簡單理解就是對稱數(shù)
解決： 每次取到第一個數(shù)和最后一個數(shù)進行對比坎缭， 對比后把首尾去掉斜友，直到原數(shù)據(jù)為0
class Solution {
func isPalindrome(_ x: Int) -> Bool {
if x < 0 {
return false
}
var div = 1
var num = x
while (num / div >= 10) {
div = div * 10
}
while(num != 0) {
let left = num / div
let right = num % 10
if left != right {
return false
}
num = (num - left * div) / 10
div = div / 100
}
return true
}
}
class Solution {
func isPalindrome(_ x: Int) -> Bool {
guard x >= 0 else {
return false
}
var num = String(x)
while(num.count > 1) {
guard let first = num.first else { return false }
guard let last = num.last else { return false }
guard first == last else { return false }
num.removeFirst()
num.removeLast()
}
return true
}
}
