LeetCode 43 - Multiply Strings

https://leetcode.com/problems/multiply-strings/

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly

https://www.cnblogs.com/yrbbest/p/4436340.html

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

int p1 = i + j, p2 = i + j + 1;
int sum = mul + pos[p2];

We only need p1 and p2 here because 9 * 9 + 9 < 100

Remember how we do multiplication?

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]` 

---

---

public String multiply(String num1, String num2) {
    int m = num1.length(), n = num2.length();
    int[] pos = new int[m + n];
   
    for(int i = m - 1; i >= 0; i--) {
        for(int j = n - 1; j >= 0; j--) {
            int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0'); 
            int p1 = i + j, p2 = i + j + 1;
            int sum = mul + pos[p2];

            pos[p1] += sum / 10;
            pos[p2] = (sum) % 10;
        }
    }  
    
    StringBuilder sb = new StringBuilder();
    for(int p : pos) if(!(sb.length() == 0 && p == 0)) sb.append(p);
    return sb.length() == 0 ? "0" : sb.toString();
}

https://blog.csdn.net/mine_song/article/details/70482898

  public String multiply(String num1, String num2) {

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

      return "0";

    int n1 = num1.length();

    int n2 = num2.length();

    int[] products = new int[n1 + n2];

    for (int i = n1 - 1; i >= 0; i--) {

      for (int j = n2 - 1; j >= 0; j--) {

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

      }

    }

    int carry = 0;

    StringBuilder sb = new StringBuilder();

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

      int tmp = products[i] + carry;

      sb.append(tmp % 10);

      carry = tmp / 10;

    }

    sb.reverse();

    if (sb.charAt(0) == '0')

      sb.deleteCharAt(0);

    return sb.toString();

  }

https://leetcode.com/problems/multiply-strings/discuss/17608/AC-solution-in-Java-with-explanation

If we break it into steps, it will have the following steps. 1. compute products from each pair of digits from num1 and num2. 2. carry each element over. 3. output the solution.

Things to note:

1. The product of two numbers cannot exceed the sum of the two lengths. (e.g. 99 * 99 cannot be five digit)

int d1 = num1.charAt(i) - '0';
int d2 = num2.charAt(j) - '0';
products[i + j + 1] += d1 * d2;

    public String multiply(String num1, String num2) {
        int n1 = num1.length(), n2 = num2.length();
        int[] products = new int[n1 + n2];
        for (int i = n1 - 1; i >= 0; i--) {
            for (int j = n2 - 1; j >= 0; j--) {
                int d1 = num1.charAt(i) - '0';
                int d2 = num2.charAt(j) - '0';
                products[i + j + 1] += d1 * d2;
            }
        }
        int carry = 0;
        for (int i = products.length - 1; i >= 0; i--) {
            int tmp = (products[i] + carry) % 10;
            carry = (products[i] + carry) / 10;
            products[i] = tmp;
        }
        StringBuilder sb = new StringBuilder();
        for (int num : products) sb.append(num);
        while (sb.length() != 0 && sb.charAt(0) == '0') sb.deleteCharAt(0);
        return sb.length() == 0 ? "0" : sb.toString();
    }




X. https://blog.csdn.net/NK_test/article/details/49623191
    string multiply(string num1, string num2) {
        string s(1000,'0');
        reverse(num1.begin(),num1.end());
        reverse(num2.begin(),num2.end());
        for(int i=0;i<num1.length();i++)
        for(int j=0;j<num2.length();j++)
        {
            int temp=(num1[i]-'0')*(num2[j]-'0');
            s[i+j+1]=s[i+j+1]-'0'+(s[i+j]-'0'+temp)/10+'0';
            s[i+j]=(s[i+j]-'0'+temp)%10+'0';
        }
        reverse(s.begin(),s.end());
        if(s.find_first_not_of('0')==string::npos)
        return "0";
        else
        return s.substr(s.find_first_not_of('0'));
    }

https://leetcode.com/problems/multiply-strings/discuss/17667/Java-reverse-stringbuilder

    public String multiply(String left, String right) {
        if(left.equals("0") || right.equals("0"))
            return "0";
        
        // use p to accumulate sum of prod.    
        int[] p = new int[left.length() + right.length()] ;
        left = new StringBuilder(left).reverse().toString();
        right = new StringBuilder(right).reverse().toString();
        
        for(int i =0; i < left.length(); i ++){
            for(int j =0; j < right.length(); j ++){
                int val = (left.charAt(i) -'0') * (right.charAt(j) -'0');
                p[i+j] += val;
            }
        }
        
        // convet p into string
        int carry =0;    
        StringBuilder sb = new StringBuilder();
        
        for(int v: p){
            v += carry;
            carry = v/10;
            v %= 10;
            
            sb.append(v);
        }                
        String ret = sb.reverse().toString();
        return ret.charAt(0) == '0' ? ret.substring(1) : ret;
    }

https://www.1point3acres.com/bbs/forum.php?mod=viewthread&tid=451849
第二轮做internal search的白人小哥哥，让输出一个数的factorial结果，其实就是批了一层皮的大卫数乘法在string上操作，很久以前做过但是写的略慢，就写了这一题。