Multiply Strings

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.

 

 1 public class Solution {

 2     public String multiply(String num1, String num2) {

 3         // Start typing your Java solution below

 4         // DO NOT write main() function

 5         String large = null;

 6         String small = null;

 7         if(num1.length() > num2.length()){

 8             large = num1;

 9             small = num2;

10         }else{

11             large = num2;

12             small = num1;

13         }

14         if(num1.equals("0") || num2.equals("0")){

15             return "0";

16         }

17         StringBuffer result = new StringBuffer();

18         int carry = 0;

19         for(int l = large.length() + small.length() - 2; l > -1; l --){

20             int sum = carry;

21             for(int j = small.length() - 1; j > -1; j --){

22                 int i = l - j;

23                 if(i < large.length() && i > -1){

24                     sum += (large.charAt(i) - '0') * (small.charAt(j) - '0');

25                 }

26             }

27             carry = sum / 10;

28             result.insert(0,sum % 10);

29         }

30         if(carry != 0){

31             result.insert(0,carry);

32         }

33         return result.toString();

34     }   

35 }

 第二遍:

利用array 可以简化运算！

 1 public class Solution {

 2     public String multiply(String num1, String num2) {

 3         // Start typing your Java solution below

 4         // DO NOT write main() function

 5         if ( num1.equals("0") || num2.equals("0") ) return "0";

 6         int n1=num1.length(), n2 = num2.length();

 7         int[] c = new int[n1 + n2];

 8         for (int i = n1 - 1; i > -1; i--){

 9             for (int j = n2 - 1; j > -1; j--){

10                 c[i + j + 1] += (num1.charAt(i)-'0')*(num2.charAt(j)-'0');

11             }

12         }

13         for (int o=n1 + n2 - 1; o > 0; o --){

14             c[o - 1] += c[o]/10;

15             c[o] %= 10;

16         }

17         int bit = 0;

18         if(c[0] == 0) bit = 1;

19         StringBuffer ret = new StringBuffer();

20         for (int i=bit; i < n1 + n2; i ++)

21             ret.append(c[i]);

22         return ret.toString();

23     }   

24 }

 第三遍：

 1 public class Solution {

 2     public String multiply(String num1, String num2) {

 3         if(num1.equals("0") || num2.equals("0")) return "0";

 4         int len1 = num1.length(), len2 = num2.length();

 5         int[] result = new int[len1 + len2 - 1];

 6         for(int i = 0; i < len1; i ++){

 7             for(int j = 0; j < len2; j ++){

 8                 result[i + j] += (num1.charAt(len1 - 1 - i) - '0') * (num2.charAt(len2 - 1 - j) - '0');

 9             }

10         }

11         int upp = 0;

12         StringBuilder sb = new StringBuilder();

13         for(int i = 0; i < result.length; i ++){

14             int sum = result[i] + upp;

15             sb.insert(0, sum % 10);

16             upp = sum / 10;

17         }

18         if(upp != 0) sb.insert(0, upp);

19         return sb.toString();

20     }

21 }