力扣7:整数反转

import java.util.*;
class Solution {
    public int reverse(int x) {
       
        int flag = x;
        if(flag < 0){
            x = -x;
        }
        int num = 0; //返回后的数字
        long lnum = 0;//用更长范围,来比较是否出错,确实比较low
        String str = Integer.toString(x);
        char arr[] = str.toCharArray();
        Reverse(arr);
        for(int i = 0; i < arr.length;i++){
            num = (arr[i]-'0') + num*10;
            lnum = (arr[i]-'0') + lnum*10;
            if(num!=lnum){
                return 0;
            }
        }
        if(flag > 0)
            return num;
        else
            return -num;
    }
    public void Reverse(char str[]){
        int len = str.length;
        char temp;
        for(int i = 0;i < len/2;i++){
            temp = str[i];
            str[i] = str[len - 1 - i];
            str[len - 1 -i] = temp;
        }
    } 
}

判断溢出比较麻烦。num做计算处try catch不知道为什么就是不行。

原文地址:https://www.cnblogs.com/theWinter/p/10592746.html