个人记录-LeetCode 6.ZigZag Conversion

问题：
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”.

其实就是将字符串按锯齿的形式进行整理后，按顺序读出来。
将人写这种字符串的方式，用代码表现一下即可。

代码示例：

public class Solution {
    public String convert(String s, int numRows) {
        if(s == null) {
            return null;
        }

        if (numRows <= 1) {
            return s;
        }

        int len = s.length();

        //为了速度比较快，采取空间换时间的方法
        //分配数组存储结果的位图
        char[][] temp = new char[numRows][len];
        for (int i = 0; i < numRows; ++i) {
            for (int j = 0; j < len; ++j) {
                temp[i][j] = '\0';
            }
        }

        char[] sChar = s.toCharArray();

        int rowIndex = 0;
        int columnIndex = 0;
        //zigFlag为true，表示在处理锯齿的斜线部分
        boolean zigFlag = false;

        for (int i = 0; i < len; ++i) {
            temp[rowIndex][columnIndex] = sChar[i];

            if (!zigFlag) {
                //正常形式，竖着写，增加行号即可
                if (rowIndex + 1 < numRows) {
                    ++rowIndex;
                } else {
                    --rowIndex;
                    ++columnIndex;

                    zigFlag = true;
                }

                continue;
            }

            //处理锯齿部分，就是斜着写，坐标处理一下就行
            if (rowIndex - 1 >= 0 && columnIndex + 1 < len) {
                --rowIndex;
                ++columnIndex;
            } else {
                zigFlag = false;
                ++rowIndex;
            }
        }

        StringBuilder sb = new StringBuilder("");
        for (int i = 0; i < numRows; ++i) {
            for (int j = 0; j < len; ++j) {
                if (temp[i][j] != '\0') {
                    sb.append(temp[i][j]);
                }
            }
        }

        return sb.toString();
    }
}