LeetCode之6 ZigZag Conversion

1.Description

The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: “PAHNAPLSIIGYIR”

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

2. TC

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P     I    N
A   L S  I G
Y A   H R
P     I

Example 3:

Input: s = "A", numRows = 1
Output: "A"

3.Solution

public String convert(String s, int numRows) {
    
      if(numRows == 1)
          return s;
       
        List<StringBuilder> ans = new ArrayList<>();
        
        for (int i = 0; i <= Math.min(numRows, s.length()); i++) {
    
            ans.add(new StringBuilder());
        }
        
        boolean goingDown = false;
        
        int curRow = 0;
        
        for(char c : s.toCharArray()) {
    
            ans.get(curRow).append(c);
            if (curRow == 0 || curRow == numRows -1)
                goingDown = !goingDown;
            curRow += goingDown ? 1 : -1;
        }
        
       StringBuilder sb = new StringBuilder();
        for(StringBuilder chars: ans){
    
                sb.append(chars);
        }
        return sb.toString();
    }

思路：
然后用一个list存放每一行的内容，每一行的内容是不连贯，因此用StringBuilder来拼接字符串。
然后根据重上倒下还是重下到上的方向来决定是在哪一行。
最后把所有StringBuilder拼接成一个完整的字符串。

public String convert(String s, int numRows) {
    
      if(numRows == 1)
          return s;
        StringBuilder sb = new StringBuilder();
        int n  = s.length();
        int df = 2*numRows -2;
        
        for (int i = 0; i < numRows; i++) {
    
            for (int j = 0; j + i < n; j+=df ) {
    
                sb.append(s.charAt(i + j));
                if (i != 0 && i != numRows - 1 && j -i +df < n) {
    
                    sb.append(s.charAt(j - i + df));
                }
            }
        }
        return sb.toString();
    }

逐行读取Z字形的内容，要算出两个点之间的差值，找到其规律。

P     I    N
A   L S  I G
Y A   H R
P     I

以上图为例，第一行的三个字符P， I， N，在原字符串中下标分别为0， 6， 12，差值为6，这个值和行数有关，3行差值为4，得出的规律是2* 行数-2；
对于中间的行数，就是j-i+df