[LeetCode#7]Reverse Integer
Reverse a integer is a very common task we need to understand for being a good programmer. It is very easy at some aspects, however, it also needs a prudent mind to understand all corner cases it may encounter.
The basic idea of reversing a number is to use mod "%" and divid "/" collectedly.
1. get the last digit of a number (ten as base, decimal numeral system)
digit = num % 10
2. chop off the last digit of a number
num_remain = num / 10
The basic idea is :
int ret = 0 while ( num_remain ! = 0 ) {
digit = num % 10;
ret = ret * 10 + digit;
num_remain = num_remain / 10; }
However, there is a big pitfall at here. The problem is that the ret_num could exceed the max number a integer could represent in Java. What's more, if the num is a negative number, the symbol of the ret_num could not be predicted. we need to add some fix for the basic structure above.
1. convert the number into positive form before reversion.
if (num < 0) { num = num * -1;
neg_flag = true; }
2. add proper mechanism to check possible overflow. (note: the ret_num is positive now!)
if (ret != 0) { if ((Integer.MAX_VALUE - digit) / ret < 10 ) return 0; if (neg_flag == true) { if (-10 < (Integer.MIN_VALUE + digit) / ret) return 0; } } ret= ret * 10 + digit;
The reason is : (when overflow happens)
iff neg_flag = fase, (note: the ret_num is positive now!)
ret * 10 + digit > Integer.MAX_VALUE <=> (Integer.MAX_VALUE - digit) / ret < 10
iff neg_flag = true,
- (ret * 10 + digit) < Integer.MIN_VALUE <=> -10 < (Integer.MIN_VALUE + digit) / ret
Questions:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
My answer:
//take care of overflow //both mod and divid can have positive/negative symbols public class Solution { //watch out for all corner cases!!! public int reverse(int x) { int ret = 0; int digit = 0; boolean neg_flag = false; if (x < 0) { neg_flag = true; x = -1 * x; //covert to abs(x), and record the symbol of negative or positive. } while (x != 0) { digit = x % 10; //get the last digit of x if (ret != 0) { //must follow this pattern to check if ((Integer.MAX_VALUE - digit) / ret < 10 ) return 0; if (neg_flag == true) { if (-10 < (Integer.MIN_VALUE + digit) / ret) // - (ret * 10 + digit) < Integer.MIN_VALUE //if we convert the number to abs, we need to compare it in negative form with Integer.MIN_VALUE return 0; } } ret = ret * 10 + digit; x = x / 10; //chop off the last digit of x } if (neg_flag == true && ret > 0) ret = -1 * ret; return ret; } }