LeetCode - Multiply Strings

Multiply Strings

2013.12.15 04:07

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

Solution:

　　I just did a digit-by-digit multiplication. There seemed to be a very complicated O(n * log(n)) solution, using Fast Fourier Transformation.

　　Here is the link to that solution, if you are eager to dip into it: http://hi.baidu.com/aekdycoin/item/0096b386bfadfd2e100ef3dc

　　Also there is an exercise problem on "Introduction to Algorithms", I'll paste my O(n * log(n)) solution here when I understand how it works (^_^)

　　Time complexity is O(n^2), where n is the length of the array. Space complexity is O(n), since storing the array requires O(n) space.

Accepted code:

 1 // 1CE, 1AC, nice performance

 2 class Solution {

 3 public:

 4     string multiply(string num1, string num2) {

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

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

 7         if(num1 == "0" || num2 == "0"){

 8             return "0";

 9         }

10         

11         int *a1, *a2, *a3;

12         int max_len, n;

13         char *buf;

14 

15         max_len = num1.length() + num2.length() + 5;

16         a1 = new int[max_len];

17         a2 = new int[max_len];

18         a3 = new int[max_len];

19         buf = new char[max_len];

20         // 1CE here, declaration for $j missing

21         int i, j;

22         

23         for(i = 0; i < max_len; ++i){

24             a1[i] = a2[i] = a3[i] = 0;

25             buf[i] = 0;

26         }

27         

28         n = num1.length();

29         for(i = 0; i <= n - 1; ++i){

30             a1[n - 1 - i] = num1[i] - '0';

31         }

32         

33         n = num2.length();

34         for(i = 0; i <= n - 1; ++i){

35             a2[n - 1 - i] = num2[i] - '0';

36         }

37         

38         for(i = 0; i < max_len; ++i){

39             if(a1[i] == 0){

40                 continue;

41             }

42             for(j = 0; i + j < max_len; ++j){

43                 a3[i + j] += a1[i] * a2[j];

44             }

45         }

46         

47         for(i = 0; i < max_len - 1; ++i){

48             a3[i + 1] += a3[i] / 10;

49             a3[i] %= 10;

50         }

51         a3[max_len - 1] %= 10;

52         i = max_len - 1;

53         while(i >=0 && a3[i] == 0){

54             --i;

55         }

56         

57         if(i < 0){

58             sprintf(buf, "0");

59         }else{

60             j = 0;

61             while(i >= 0){

62                 buf[j++] = a3[i--] + '0';

63             }

64         }

65         

66         string res = string(buf);

67         

68         delete[] a1;

69         delete[] a2;

70         delete[] a3;

71         delete[] buf;

72         return res;

73     }

74 };