[TOC]
P006 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".
思路分析
5行
6行
最终结果就是将上图中每一行的内容去掉中间间隔连接起来
所以最直观的方法就是：
- 生成类似上图的结构(二维数组？？？)
- 去掉间隔
- 连接每一行
另外：
若使用二维数组，最终连接的时候还得去掉空白。所以可以使用特殊的二维数组--一维字符串数组来保存上述结构的同时去掉间隔。
代码
java
import java.util.Arrays;
public class Solution006 {
public String convert(String s, int numRows) {
if (s == null || s.length() == 0 || numRows <= 1)
return s;
String rows[] = new String[numRows];
Arrays.fill(rows, "");
boolean down = true;
int row = 0;
for (int i = 0; i < s.length(); i++) {
rows[row] += s.charAt(i);
if (down) {
row++;
} else {
row--;
}
if (row >= numRows) {
//不是减一，因为第一行行最后一行都是一个元素
row = numRows - 2;
down = false;
}
if (row < 0) {
//不是零,因为第一行行最后一行都是一个元素
row = 1;
down = true;
}
}
StringBuilder sb = new StringBuilder();
for (String r : rows) {
sb.append(r);
}
return sb.toString();
}
}
python
class Solution006(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
if not s or len(s) == 0 or numRows <= 1:return s
rows = [""] * numRows
down = True;row = 0;
for e in s:
rows[row] += e
if down:
row += 1
else:
row -= 1
if row >= numRows:
row = numRows - 2
down = False
if row < 0:
row = 1
down = True
# end for
ret = ""
for e in rows:
ret += e
return ret
