Leetcode-43- string multiplication

subject

Ideas

1. 1 <= num1.length, num2.length <= 200, That means there may be 200 Multiplication of digits , That is, there is currently no data type that can store these numbers , Therefore, it cannot be converted into numbers and multiplied directly .
2. It's good if you add it up , Record whether to carry , Just a few of you , But the multiplication is a little troublesome …
3. Painstakingly finished , Think I'm retarded , Let's go and have a look , I feel that the code of the problem solution is much simpler than mine , But it is also done by vertical thinking .

Code

    public String multiply(String num1, String num2) {
    
        if(num1.equals("0") || num2.equals("0")) return "0";
        LinkedList<String> ll = new LinkedList<>();
        // Record the number multiplied by each round   hold num1 Put it on top.  num2 Put it underneath. 
        for(int i=num2.length()-1;i>=0;i--){
    
                ll.add(mm(num1,num2.charAt(i)));
        }
        // After recording   Start adding up the numbers 
        String ans = calute(ll);
        return ans;

    }
    // Record a long string   The result of multiplying a short string with only one bit 
    public String mm(String num1,Character c){
    
        StringBuilder sb = new StringBuilder();
        int left = (int)c-'0';
        // carry 
        int flag=0;
        for(int i=num1.length()-1;i>=0;i--){
    
            Character temp = num1.charAt(i);
            int right = (int)temp - '0';
            // Turn to numbers 
            int ans = left*right+flag;
            // If at this time i For the last one   Then you can directly reverse and add 
            if(i==0){
    
                // If ans by 20 30  Then only 2 3 So we have to deal with it separately 
                if(ans>=10){
    
                    StringBuilder ans_str = new StringBuilder(String.valueOf(ans));
                    sb.append(ans_str.reverse().toString());
                }else sb.append(ans);
            }
            else {
    
                if(ans>=10){
    
                    flag = ans / 10;
                    ans = ans % 10;
                }else {
    
                    flag=0;
                }
                sb.append(ans);
            }
        }
        return sb.reverse().toString();
    }
    // Get the final result 
    public String calute(LinkedList<String> ll){
    
        int index=0;
        // Add a few zeros to the right 
        int right = ll.size()-1;
        StringBuilder sb = new StringBuilder();
        // To expand each bit in the linked list to the current length 
        // You can add while expanding   Record the final answer 
        String ans="";
        for(int i=ll.size()-1;i>=0;i--){
    
            // Take out the string 
            String temp = ll.get(i);
            // Add a few zeros to the left 
            int left=0;
            if(ans.equals("")) left = ll.get(ll.size()-1).length()-temp.length();
            else left = ans.length()-temp.length()-right;
            for(int j=0;j<left;j++) sb.append("0");
            sb.append(temp);
            for(int j=0;j<right;j++) sb.append("0");
            if(ans=="") ans = sb.toString();
            else {
     // take ans And sb.toString() Add up 
                ans = ad(ans,sb.toString());
            }
// ans_array[index++] = sb.toString();
            sb.delete(0, sb.length());
            right--;
        }
        return ans;
    }
    // Calculate two numbers of the same number String Add up 
    public String ad(String nums1,String nums2){
    
        StringBuilder sb = new StringBuilder();
        // carry 
        int flag=0;
        for(int i=nums1.length()-1;i>=0;i--){
    
            int left = (int)nums1.charAt(i)-'0';
            int right = (int)nums2.charAt(i)-'0';
            int ans = left+right+flag;
            if(i==0){
    
                // If ans by 20 30  Then only 2 3 So we have to deal with it separately 
                if(ans>=10){
    
                    StringBuilder ans_str = new StringBuilder(String.valueOf(ans));
                    sb.append(ans_str.reverse().toString());
                }else sb.append(ans);
            }
            else {
    
                if(ans>=10){
    
                    flag = ans / 10;
                    ans = ans % 10;
                }else {
    
                    flag=0;
                }
                sb.append(ans);
            }
        }
        return sb.reverse().toString();
    }