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".

 

分析：

　　这道题主要是找规律。

方法一：

　　比较直观的解法，使用list存储每一行，最后拼接。

 

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        res = ['']*numRows
        step = 1
        row = 0
        if len(s) <= 2 or numRows <= 1:
            return s
        for i in range(len(s)):
            res[row] += s[i]
            row += step
            if row >=numRows:
                row = numRows-2
                step = -1
            if row < 0:
                row = 1
                step = 1
        ans = ''
        for i in range(len(res)):
            ans += res[i]
        return ans

 

 

 

方法二：

　　发现所有行的重复周期都是 2 * nRows - 2，对于首行和末行之间的行，还会额外重复一次，重复的这一次距离本周期起始字符的距离是 2 * nRows - 2 - 2 * i

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        if len(s) < 2 or numRows < 2:
            return s
        jump = 2*(numRows-1)
        res = ''
        for i in range(numRows):
            j = i
            while j < len(s):
                res += s[j]
                if i > 0 and i < numRows-1 and j+jump-2*i < len(s):
                    res += s[j+jump-2*i]
                j += jump
        return res

 

转载于:https://www.cnblogs.com/Peyton-Li/p/7648549.html