整数反转

又是一次失败的Debug型编程呢_(:з」∠)_

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

Example

Input: 123
Output: 321

Input: -123
Output: -321

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.

我的（ 3 ms ）

public int reverse(int x) {
		if (x > 2147483647||x <= -2147483648){
			return 0;//防止输入溢出
		}
		int y =Math.abs(x);//防止负数
		String str   = String.valueOf(y);   
		int [] tmp 	 = new int[str.length()];
		for(int i=0;i<str.length();i++){
		tmp[i] = Integer.parseInt(String.valueOf(str.charAt(i)));
		}

		long sum=0;//防止输出溢出

		for(int j=tmp.length-1;j>=0;j--){
    	sum=sum*10+tmp[j];
		}

		if (x<0){
			sum= 0-sum ;//恢复负数
		}
		if (sum > 2147483647||sum < -2147483648){
			return 0;//防止输出溢出

		}
		
		return (int)sum;

    }

先把数字转换成数组，然后再把数组转换回数字，中间处理好负数和溢出的问题。

1 ms

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while(x != 0){
            int pop = x % 10;
            x = x / 10;
            if(rev > Integer.MAX_VALUE / 10 || (rev == Integer.MAX_VALUE / 10 && pop > Integer.MAX_VALUE % 10)){
                rev = 0;
                break;
            }else if(rev < Integer.MIN_VALUE / 10 || (rev == Integer.MIN_VALUE / 10 && x < Integer.MIN_VALUE % 10)){
                rev = 0;
                break;
            }
            rev = rev * 10 + pop;
        }
        return rev;
    }
}

2 ms （谁都不服就服你！）

class Solution {
    public int reverse(int x) {
       long n = 0;
       while(x!=0){
           n = n*10 + x%10;
           x/=10;
       }
       return (int)n==n? (int)n:0;//牛逼！！
    }
}