Force buckle 6 Zigzag transformation

Title source ：https://leetcode-cn.com/problems/zigzag-conversion/

General meaning ：
Give a string and number of lines r, String from top to bottom 、 Left to right Z Font arrangement , Then, the arranged strings are sorted from left to right 、 Take them out from top to bottom 、

Ideas

Directly simulate the transformation process

According to the title , stay Z In the process of transformation , I will fill it down first r Characters , And then fill it on the right r - 2 Characters , This process can be regarded as a cycle t = r * 2 - 2, So there are n / t （ Rounding up , The last cycle may not be complete , But it doesn't affect ） A cycle

After calculating the period , It can be simulated according to the periodic change rules , To save memory , You can use it directly StringBuffer Represents the string corresponding to each line

simulation

1. Find the period t
2. Traversing the entire string , The index for i. Set the line mark at the beginning x = 0. So when i mod t < r - 1 when , You need to continue to fill in characters , Row number x + 1; Otherwise, fill in the characters to the right , Row number x - 1
3. Index each time i Place the character at in the line marker x Corresponding StringBuffer End of object , Then based on the current index i Determine the number of rows that should be placed next time

Code ：

public String convert(String s, int numRows) {
    
        int n = s.length();
        int r = numRows;
        //  If the number of rows is  1  Or greater than the string length , Directly return the original string 
        if (r == 1 || r >= n) {
    
            return s;
        }
        //  cycle 
        int t = r * 2 - 2;
        //  The string object corresponding to the row 
        StringBuffer[] sb = new StringBuffer[r];
        for (int i = 0; i < r; i++) {
    
            sb[i] = new StringBuffer();
        }
        //  Traverse 
        for (int i = 0, x = 0; i < n; i++) {
    
            //  Put the current character on the corresponding line 
            sb[x].append(s.charAt(i));
            //  Determine the number of lines to put characters next time 
            if (i % t < r - 1) {
    
                x++;
            } else {
    
                x--;
            }
        }
        StringBuffer ans = new StringBuffer();
        for (int i = 0; i < r; i++) {
    
            ans.append(sb[i]);
        }
        return ans.toString();
    }