LeetCode: ZigZag Conversion 解题报告

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 text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

SOLUTION 1:

使用以下算法会比较简单。

两个规律：

1 两个zigzag之间间距为2*nRows-2

2 每个zigzag中间（在j和j+interval之间）位置为j+interval-2*i

注意：当Rows = 1时，此方法不适用，因为size = 0,会造成死循环。所以Rows = 1时，需要独立处理。

引自：http://blog.csdn.net/fightforyourdream/article/details/16881517

public class Solution {
    public String convert(String s, int nRows) {
        if (s == null) {
            return null;
        }
        
        // 第一个小部分的大小        
        int size = 2 * nRows - 2;
        
        // 当行数为1的时候，不需要折叠。
        if (nRows <= 1) {
            return s;
        }
        
        StringBuilder ret = new StringBuilder();
        
        int len = s.length();
        for (int i = 0; i < nRows; i++) {
            // j代表第几个BLOCK
            for (int j = i; j < len; j += size) {
                ret.append(s.charAt(j));
                
                // 即不是第一行，也不是最后一行，还需要加上中间的节点
                int mid = j + size - i * 2;
                if (i != 0 && i != nRows - 1 && mid < len) {
                    char c = s.charAt(mid);
                    ret.append(c);
                }
            }
        }
        
        return ret.toString();
    }
}

 2015.1.4 redo:

public class Solution {
    public String convert(String s, int nRows) {
        if (s == null) {
            return null;
        }
        
        // corner case;
        if (nRows == 1) {
            return s;
        }
        
        // The number of elements in a section.
        int section = 2 * nRows - 2;
        StringBuilder sb = new StringBuilder();
        
        int len = s.length();
        for (int i = 0; i < nRows; i++) {
            for (int j = i; j < len; j += section) {
                char c = s.charAt(j);
                sb.append(c);
                
                // The middle rows.
                int mid = j + section - 2 * i;
                // bug 2: the mid is out of range.
                if (i != 0 && i != nRows - 1 && mid < len) {
                    // bug 1: forget a ')'
                    sb.append(s.charAt(mid));
                }
            }
        }
        
        return sb.toString();
    }
}

 

请至主页君的GIT HUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/Convert.java