LeetCode第7題_Reverse Integer

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.

思路:先嘗試取到各個字的digit，在反序去進行打印

改寫為題目對應的函式

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public static int reverse(int x){ ArrayList revRes = new ArrayList(); int src = x; if(x<0){ x = x*(-1); } while(x>0){ int digit = x%10; revRes.add(digit); x/=10; } int res = Integer.parseInt(revRes.toString().replace("[", "").replace("]", "").replace(", ", "")); if(src<0){ res = res * (-1); return res; } return res; }

要注意數值範圍越界溢位的問題

第二版.(過關版)

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/* * To change this license header, choose License Headers in Project Properties. * To change this template file, choose Tools | Templates * and open the template in the editor. */ package leetcode7_reverseinteger; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; /** * * @author chous */ public class LeetCode7_ReverseInteger { /** * @param args the command line arguments */ public static void main(String[] args) throws IOException { // TODO code application logic here BufferedReader input; input = new BufferedReader(new InputStreamReader(System.in)); //System.out.println("Integer.MAX:" + Integer.MAX_VALUE); //System.out.println("Integer.Min:" + Integer.MIN_VALUE); System.out.println("Input"); String strInput = input.readLine(); int numInput = Integer.parseInt(strInput); System.out.println(numInput); System.out.println(reverse(numInput)); } public static int reverse(int x){ int temp = 0; int result = 0; while (x != 0) { temp = result * 10 + x % 10; // 判断是否越界 if(temp / 10 != result) { return 0; } x = x / 10; result = temp; } return result; } }