leetcode刷题笔记七 整数反转 Scala版本

leetcode刷题笔记七 整数反转 Scala版本

源地址：(leetcode刷题笔记七 整数反转 Scala版本)[https://leetcode.com/problems/reverse-integer/submissions/]

问题描述：

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

简要思路分析：

对于整数反转，可以从字符串或者Long型数据处理。对于字符串处理方法，我们需要将符号与数字剥离，通过对原符号推出转换后的符号，对气候的数字串进行反转，将首部的0处理掉；对于Long型数据处理，针对X不断取余和除10 将个位数分离出来作为转换后的数的高位处理。这两种都要解决的问题是Int型数反转后溢出的情况进行处理，这是本题的关键！

代码补充：

法一：

object Solution {
def reverse(x: Int): Int = {
        var xStr = x.toString
        var Xbol = ""
        if (xStr.length == 0 || xStr.length == 1) return xStr.toInt
        if (xStr.charAt(0) == '-') {
          Xbol = "-"
          xStr = xStr.substring(1).reverse
        }
        else xStr = xStr.reverse
        while(xStr.charAt(0) == '0'){
          xStr = xStr.substring(1)
        }
        //println(Xbol.concat(xStr))
        val x1 = (Xbol.concat(xStr)).toLong
        if (x1 > Int.MaxValue || x1 < Int.MinValue) {
            return 0
        }
        else
        return x1.toInt
    }
}

法二：

object Solution {
def reverse(x: Int): Int = {

      if (x > Int.MaxValue || x < Int.MinValue) {
        return 0
      }

      var xTemp = x
      var result:Long = 0
      while (xTemp != 0){
        result = result * 10 + xTemp % 10
        xTemp /= 10
      }

      if (result > Int.MaxValue || result < Int.MinValue) {
        return 0
      }
      else
        return  result.toInt

    }
}