LeetCode - Divide Two Integers

Divide Two Integers

2013.12.7 04:52

Divide two integers without using multiplication, division and mod operator.

Solution:

　　Another bit-manipulation problem, integer division. Normally if we calculate x / y (x, y > 0), we'll subtract y from x until y < x. This operation can be accelerated if we subtract x << (..., 2, 1, 0) from y, until x < y, using only log2(x) time.

　　Time complexity is O(log(dividend) - log(divisor)), space complexity is O(1).

Accepted code:

 1 // 2RE, 2WA, 1AC

 2 class Solution {

 3 public:

 4     int divide(int dividend, int divisor) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.

 7         // 2RE here, -2147483648 is a tricky case

 8         

 9         long long int div1, div2;

10         int s1, s2;

11         long long int base;

12         // 1WA here, $r overflows, must be long long int

13         long long int res, r;

14         

15         s1 = sign(dividend);

16         s2 = sign(divisor);

17         

18         if(s1 * s2 == 0){

19             return 0;

20         }

21         

22         div1 = dividend > 0 ? dividend : -(long long int)dividend;

23         div2 = divisor > 0 ? divisor : -(long long int)divisor;

24         

25         base = div2;

26         r = 1;

27         while(base <= div1){

28             base <<= 1;

29             r <<= 1;

30         }

31         

32         res = 0;

33         while(r > 0){

34             if(div1 >= base){

35                 div1 -= base;

36                 res += r;

37             }

38             base >>= 1;

39             r >>= 1;

40         }

41         

42         if(s1 < 0){

43             res = -res;

44         }

45         if(s2 < 0){

46             res = -res;

47         }

48         

49         // 1WA here, forgot to multiply sign variables

50         return res;

51     }

52 private:

53     int sign(int x){

54         if(x > 0){

55             return 1;

56         }else if(x < 0){

57             return -1;

58         }else{

59             return 0;

60         }

61     }

62 };