6. Z 字形变换 ZigZag Conversion

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);

 

方法一：

    从左向右迭代字符串。

    

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

时间复杂度O（n）

 

方法二：

    字符从第一行到最后一行再回到第一行的周期为2*numRows−2

    第一行字符位于原来的 k  (2*numRows−2)

    最后一行字符位于原来k(2*numRows−2)+numRows−1   

    中间第i行字符位于原来的k(2*numRows−2)+i 和(k+1)(2*numRows−2)−i

    

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

时间复杂度O(n)

 

参考链接：

https://leetcode.com/problems/zigzag-conversion

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