LeetCode：ZigZag Conversion

题目链接：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)

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”.

第一版代码，通过模拟ZigZag Conversion的过程得到一个字符矩阵，然后遍历矩阵得到结果，代码比较丑，然而我也不想改了，将就看吧：

public class Solution {    public String convert(String s, int numRows) {        if (s == null || numRows == 1) {            return s;        }        int length = s.length();        int baseRows = numRows - 1;        // 计算结果矩阵的列数        int numColumns = (length / (numRows + numRows-2)) * baseRows;        int remainder = length % (numRows + numRows-2);        if (remainder > numRows) {            numColumns += 1+remainder-numRows;        } else if (0 < remainder && remainder <= numRows) {            numColumns++;        }        char[][] temp = new char[numRows][numColumns];        // 模拟 ZigZag Conversion 过程，构造字符矩阵        for (int j = 0, index = 0; j < numColumns; j++) {            if ((j % baseRows) == 0) {                for (int i = 0; i < numRows && index < length; i++,index++) {                    temp[i][j] = s.charAt(index);                }            } else {                for (int i = numRows-2; i > 0 && index < length; i--,j++,index++) {                    temp[i][j] = s.charAt(index);                }                j--;            }        }        // 先行后列遍历矩阵，取出结果        StringBuilder val = new StringBuilder();        for (int i = 0; i < numRows; i++) {            for (int j = 0; j < numColumns; j++) {                if (temp[i][j] != 0) {                    val.append(temp[i][j]);                }            }        }        return val.toString();    }}

第二版代码，直接计算字符出现的顺序，不构造矩阵：

public class Solution {    public String convert(String s, int numRows) {        if (s == null || numRows == 1) {            return s;        }        int len = s.length();        StringBuilder val = new StringBuilder();        // 逐行添加字符        for (int i=0; i<numRows; i++) {            // 这里为了区分下降和上升两种情况，设置了一个布尔变量来进行控制            boolean flag = true;            for (int j=i; j<len; flag=!flag) {  // 下降和上升总是交替的                int pos = j;                if (flag) {                         j += (numRows-i-1)*2;   // 接下来是下降，很显然要填满下面的行（画一个凹曲线）                } else {                    j += i*2;               // 接下来是上升，很显然要填满上面的行（画一个凸曲线）                }                if (j == pos) {             // 当前值处在矩阵边界                    continue;                }                val.append(s.charAt(pos));            }        }        return val.toString();    }}

更爽的代码高亮，请看作业部落黑色主题