[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

“.

https://leetcode.com/problems/zigzag-conversion/

n=2时，字符串坐标变成zigzag的走法就是：

0 2 4 6

1 3 5 7

n=3时的走法是：

0     4     8

1  3 5  7 9

2     6    10

n=4时的走法是：

0      6        12

1   5 7    11 13

2 4   8 10    14

3      9         15

观察坐标变化 第一行和最后一行，相邻行的坐标差都是

2n-2

，其他行坐标

j+(2n-2)-2i

,以行遍历

class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
if (len(s)==0 or numRows <=0):
return ''
if numRows == 1:
return s
ret = []
size = 2*numRows-2
for i in range(0,numRows):
for j in range(i,len(s),size):
ret.append(s[j])
if (i!=0 and i!=numRows-1):
temp = j + size -2*i
if temp<len(s):
ret.append(s[temp])
return ''.join(ret)

既然用python做，那么应该很少代码才对，利用

list[-1]

list[-2]

的特性

ABCDEFG

A   E
BD
C   F
# n=3 将空格出压缩到下面形式
AE
BDG
CF

换个形式看上面的”压缩”后的结果

result[1]=

AE

result[2]=

BDG

result[3]=

CF

然后

AE

+

BDG

+

CF

，这样就得到最后结果了。～

class Solution(object):
def convert(self, s, numRows):
result = [""] * (numRows + 1) #保存每一行的结果
level, order = 1, 1 # order指示是否逆序
for i in s:
result[level*order] += i #对应行加入字母
level += 1
if level >= numRows:   # 判断是否应该逆序
level, order = 1, order * (-1)
return "".join(result)