2013.12.1 22:12
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).
Solution:
Reversing an integer can be done with a while-loop, one digit at a time. These two cases need special handling:
1. For '-123', you don't want '321-' as the answer, right?
2. For 2147483647, the reverse overflows the int32 limit, i.e. INT_MAX. How would you process such cases? Your call.
Time complexity is O(lg(n)), space complexity is O(1).
Accepted code:
1 class Solution { 2 public: 3 int reverse(int x) { 4 // Note: The Solution object is instantiated only once and is reused by each test case. 5 long long int res, lx; 6 7 if(x < 0){ 8 return -reverse(-x); 9 } 10 11 lx = x; 12 res = 0; 13 while(lx > 0){ 14 res = res * 10 + lx % 10; 15 lx /= 10; 16 } 17 18 return res; 19 } 20 };
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