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();
}
}