解题思路

字符串反转

 public int reverse(int x) {
        if (x == Integer.MIN_VALUE) { //如果不做这个判断，下面的x=-x将会报错
            return 0;
        }
        if(x <0)
            return -reverse(-x);
        String number = String.valueOf(x);
        char[] arr = number.toCharArray();
        int n = arr.length;
        for(int i =0;i<n/2;i++){
            char temp = arr[i];
            arr[i] = arr[n-1-i];
            arr[n-1-i]=temp;
        }
        String res = String.valueOf(arr);
        long numRes = Long.valueOf(res);
        return  numRes <= Integer.MAX_VALUE ? (int)numRes : 0;  //判断溢出
}

数字反转

 public int reverse(int x) {
        if (x == Integer.MIN_VALUE) { //如果不做这个判断，下面的x=-x将会报错
            return 0;
        }
        if(x <0)
            return -reverse(-x);
        int res = 0;
        while(x!=0){
            if(res>Integer.MAX_VALUE/10)
                return 0;
            res = res *10+x%10;
            x/=10;
        }
        return  res <= Integer.MAX_VALUE ? res : 0;
    }