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.
• Converting the input string to integer is NOT allowed.
• You should NOT use internal library such as BigInteger.

分析：

把结果放在一个数组里，用offset来移动起始位置。

 1 public class Solution {
 2     public String multiply(String num1, String num2) {
 3         if (num1 == null || num1.length() == 0 || num2 == null || num2.length() == 0) return "";
 4         int[] result = new int[num1.length() + num2.length()];  // maximum length
 5         
 6         int offset = 0;
 7         for (int i = num1.length() - 1; i >= 0; i--) {
 8             int number1 = num1.charAt(i) - '0';
 9             int carry = 0;
10             for (int j = num2.length() - 1; j >= 0; j--) {
11                 int number2 = num2.charAt(j) - '0';
12                 int value = number1 * number2 + result[(result.length - 1) - offset + j - (num2.length() - 1)] + carry;  
13                 carry = value / 10;  
14                 value = value % 10;  
15                 result[(result.length - 1) - offset + j - (num2.length() - 1)] = value;                  
16             }  
17             result[(result.length - 1) - offset - 1 - (num2.length() - 1)] = carry;  
18             offset++;  
19         }  
20         String str = "";  
21         boolean isBeginningZero = true;  
22         for (int i = 0; i < result.length; i++) {  
23             if (!(result[i] == 0 && isBeginningZero == true)) {
24                 isBeginningZero = false;  
25                 str += result[i];  
26             }  
27         }  
28         return str == "" ? "0" : str;  
29     }
30 }

 方法二：

https://leetcode.com/problems/multiply-strings/discuss/17605/Easiest-JAVA-Solution-with-Graph-Explanation

Start from right to left, perform multiplication on every pair of digits, and add them together. Let's draw the process! From the following draft, we can immediately conclude:

 `num1[i] * num2[j]` will be placed at indices `[i + j`, `i + j + 1]` 

---

 

 

---

 1 public class Solution {
 2     public String multiply(String num1, String num2) {
 3         int m = num1.length(), n = num2.length();
 4         int[] pos = new int[m + n];
 5 
 6         for (int i = m - 1; i >= 0; i--) {
 7             for (int j = n - 1; j >= 0; j--) {
 8                 int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
 9                 int p1 = i + j, p2 = i + j + 1;
10                 int sum = mul + pos[p2];
11                 pos[p1] += sum / 10;
12                 pos[p2] = sum % 10;
13             }
14         }
15 
16         StringBuilder sb = new StringBuilder();
17         for (int p : pos) {
18             if (!(sb.length() == 0 && p == 0)) {
19                 sb.append(p);
20             }
21         }
22         return sb.length() == 0 ? "0" : sb.toString();
23     }
24 }

 

转载于:https://www.cnblogs.com/beiyeqingteng/p/5727811.html