public class P0007 {

public int reverse(int x) { // x = 421 O(logN)+O(1)

double Max = Math.pow(2, 31) - 1; // 123

double Min = -Math.pow(2, 31); // -124

//int Max = Integer.MAX_VALUE;

//int Min = Integer.MIN_VALUE;

int res = 0; // 0 -> 1 -> 12 -> overflow

while (x != 0) { // x = 4

int remain = x % 10; // 4

if (res > Max / 10 || (res == Max / 10 && remain > Max % 10)) {

return 0;

}

if (res < Min / 10 || (res == Min / 10 && remain < Min % 10)) {

return 0;

}

res = res * 10 + remain; // 12 * 10 + 4 = 124 > 123

x /= 10;

}

return res;

}

public int reverse1(int x){

int res=0;

while(x!=0){

int remain=x%10;

int newRes=res*10+remain;//假设一个新的数来试探是否越界

if(newRes/10!=res){

return 0;

}

res=newRes;

x/=10;

}

return res;

}

}

//两个if语句提前判定下一步的倒转数是否有溢出的可能性，如果没有则继续*10+余数

// P0007 p7=new P0007();

// System.out.println(p7.reverse(123));

// System.out.println(p7.reverse(-123));

// System.out.println(p7.reverse(120));

// System.out.println(p7.reverse(0));