LeetCode 6 -- 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".

---

自己一上来没搞懂什么是zigzag，还以为就是多了个中点，结果怎么都不对，搞得相当复杂，权且贴上代码，当个教训吧：

public class ZigzagConversion {
    public static String convert(String s, int numRows) {
        if (s == null || s.length() == 0)
            return s;
        if (numRows < 2)
            return s;
        int gap = numRows - 2;
        String res = "";

        for (int i = 0; i < numRows; i++) {
            if (i == 0) {
                for (int j = i; j < s.length(); j += (numRows * 2 - 2))
                    res += s.charAt(j);
            } else if (i == numRows - 1) {
                for (int j = i; j < s.length(); j += (numRows * 2 - 2))
                    res += s.charAt(j);
            } else {
                for (int j = i; j < s.length(); j += (numRows * 2 - 2)) {
                    res += s.charAt(j);
                    if (j + numRows * 2 - 3 - i < s.length())
                        res += s.charAt(j + numRows * 2 - 3 - i);
                }
            }
        }

        return res;
    }

    public static void main(String[] args) {
        String s = "ABCDE";
        String res = convert(s, 4);
        System.out.println(res);
    }

}

问题代码，边界问题没有解决好

public class Solution {
    public String convert(String s, int nRows) {
        if (s == null || s.length() <= nRows || nRows <= 1) return s;
        StringBuffer sb = new StringBuffer();
        // the first row
        for (int i = 0; i < s.length(); i += (nRows - 1) * 2) {
            sb.append(s.charAt(i));
        }
        
        for (int i = 1; i < nRows - 1; i++) {
            for (int j = i; j < s.length(); j+= (nRows - 1) * 2) {
                sb.append(s.charAt(j));
                if (j + (nRows - i - 1) * 2 < s.length()) {
                    sb.append(s.charAt(j + (nRows - i - 1) * 2));
                }
            }
        }
        // the last row
        for (int i = nRows - 1; i < s.length(); i += (nRows - 1) * 2) {
            sb.append(s.charAt(i));
        }
        return sb.toString();
    }
}

后来百度了下，看到别人代码相当简洁，方知自己压根就没搞懂题目。

Zigzag:即循环对角线结构（

0				8				16
1			7	9			15	17
2		6		10		14		18
3	5			11	13			19
4				12				20

）

构建n个字符串，循环为每个分别添加相应的字符，最后合并所有字符；

自己照着思路重新写了下：

public class ZigzagConversion {
    public static String convert(String s, int numRows) {
        if (s == null || s.length() == 0)
            return s;
        if (numRows < 2)
            return s;
        String[] res = new String[numRows];
        String result = "";
        
        int i = 0, j;

        while (i < s.length()) {
            for (j = 0; j < numRows && i < s.length(); j++) {
                res[j] += s.charAt(i);
                i++;
            }
            for (j = numRows - 2; j > 0 && i < s.length(); j--) {
                res[j] += s.charAt(i);
                i++;
            }
        }
        for(int k = 0; k < numRows;k++)
            result += res[k];
        return result;
    }

    public static void main(String[] args) {
        String s = "PAYPALISHIRING";
        String res = convert(s, 3);
        System.out.println(res);
    }
}

只是没次字符串数组都默认初始化为null，如何覆盖还没解决，原文用C++代码如下：

string convert(string s, int nRows){
    if(nRows == 1) return s;
    string res[nRows];
    int i = 0, j, gap = nRows-2;
    while(i < s.size()){
        for(j = 0; i < s.size() && j < nRows; ++j) res[j] += s[i++];
        for(j = gap; i < s.size() && j > 0; --j) res[j] += s[i++];
    }
    string str = "";
    for(i = 0; i < nRows; ++i)
        str += res[i];
    return str;
}

发现自己一些问题：首先是不能很好的思考问题，或者说自己的思路多少总会有些偏差；其次就是边界问题，总是犯些小错误。