1+ package easy ;
2+ /**7. Reverse Integer QuestionEditorial Solution My Submissions
3+ Total Accepted: 155938
4+ Total Submissions: 655150
5+ Difficulty: Easy
6+ Reverse digits of an integer.
7+
8+ Example1: x = 123, return 321
9+ Example2: x = -123, return -321*/
10+ public class ReverseInteger {
11+ //lastly, after making it AC'ed on my own, I turned to Discuss, and find this very short solution,
12+ //it turns out we don't need to do any handling for the sign
13+ public int reverse_short_version (int x ){
14+ long rev =0 ;
15+ while (x !=0 ){
16+ rev =rev *10 +x %10 ;
17+ x /=10 ;
18+ if (rev >Integer .MAX_VALUE ||rev <Integer .MIN_VALUE )return 0 ;
19+ }
20+ return (int )rev ;
21+ }
22+
23+ //made it AC'ed on my own, cheers!
24+ public int reverse (int x ) {
25+ if (x ==0 )return 0 ;
26+ //save the first bit if it's a negative sign
27+ StringBuilder sb =new StringBuilder ();
28+ sb .append (x );
29+ boolean negative =sb .toString ().charAt (0 ) =='-' ?true :false ;
30+ //use modulor and division and use long as the result type to avoid overflow
31+ long longX = (long )x ;
32+ if (negative ){
33+ //get rid of the first '-' bit
34+ String withoutNegativeSign =sb .substring (1 ).toString ();
35+ longX =Long .parseLong (withoutNegativeSign );
36+ }
37+ sb .setLength (0 );
38+ long result =0 ;
39+ if (negative ){
40+ sb .append ('-' );
41+ }
42+ while (longX !=0 ){
43+ sb .append (longX %10 );
44+ longX /=10 ;
45+ }
46+ result =Long .parseLong (sb .toString ());
47+ System .out .println (result );//it's right here, but after converting it into an int, it overflowed to become a wrong number, how to handle this?
48+ //it turns out depending on the question requirement, on this OJ, it's expecting 0, if it's beyond the range of (Integer.MIN_VALUE, Integer.MAX_VALUE)
49+ return (result >Integer .MAX_VALUE ||result <Integer .MIN_VALUE ) ?0 : (int )result ;
50+ }
51+
52+ public static void main (String ...strings ){
53+ ReverseInteger test =new ReverseInteger ();
54+ //when the input is 1534236469, it's expecting 0 as the correct answer, this is due to its reversed number is greater than Integer.MAX_VALUE, thus return 0
55+ System .out .println (1534236469 >Integer .MAX_VALUE );
56+ System .out .println ("1534236469\n " +Integer .MAX_VALUE );
57+ // System.out.println(test.reverse(-2147483648));
58+ }
59+
60+ // this is not going to work when the input number's reverse version is greater than
61+ // Integer.MAX_VALUE, it'll throw NumberFormatException as Java cannot handle it, overflowed.
62+ public int reverse_overflowed (int x ) {
63+ //save the first bit if it's a negative sign
64+ StringBuilder sb =new StringBuilder ();
65+ sb .append (x );
66+ boolean negative =sb .toString ().charAt (0 ) =='-' ?true :false ;
67+ char []bits =sb .toString ().toCharArray ();
68+ sb .setLength (0 );
69+ if (negative ){
70+ sb .append ('-' );
71+ //until i > 0
72+ for (int i =bits .length -1 ;i >0 ;i --){
73+ sb .append (bits [i ]);
74+ }
75+ }else {
76+ //until i >= 0
77+ for (int i =bits .length -1 ;i >=0 ;i --){
78+ sb .append (bits [i ]);
79+ }
80+ }
81+ return Integer .parseInt (sb .toString ());
82+ }
83+
84+ }