6. ZigZag Conversion

【题目】

 

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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

 

 

1） 比较直观的解法是，用一个字符串数组 string[rows] 来存储每一行，最后一拼接就是最终结果。

用一个delta表示正向还是反向，即上图中从第一行到最后一行还是最后一行到第一行

public class Solution {  
    public String convert(String s, int nRows) {  
        int len = s.length();  
        if (len == 0 || nRows <= 1) return s;  
          
        String[] ans = new String[nRows];  
        Arrays.fill(ans, "");  
        int row = 0, delta = 1;  
        for (int i = 0; i < len; i++) {  
            ans[row] += s.charAt(i);  
            row += delta;  
            if (row >= nRows) {  
                row = nRows-2;  
                delta = -1;  
            }  
            if (row < 0) {  
                row = 1;  
                delta = 1;  
            }  
        }  
          
        String ret = "";  
        for (int i = 0; i < nRows; i++) {  
            ret += ans[i];  
        }  
        return ret;  
    }  
}  

 

2） 发现所有行的重复周期都是 2 * nRows - 2；

　　对于首行和末行之间的行，还会额外重复一次，重复的这一次距离本周期起始字符的距离是 2 * nRows - 2 - 2 * i

public class Solution {  
    public String convert(String s, int nRows) {  
        int len = s.length();  
        if (len == 0 || nRows < 2) return s;  
          
        String ret = "";  
        int lag = 2*nRows - 2; //循环周期  
        for (int i = 0; i < nRows; i++) {  
            for (int j = i; j < len; j += lag) {  
                ret += s.charAt(j);  
                  
                //非首行和末行时还要加一个  
                if (i > 0 && i < nRows-1) {  
                    int t = j + lag - 2*i;  
                    if (t < len) {  
                        ret += s.charAt(t);  
                    }  
                }  
            }  
        }  
        return ret;  
    }  
}