6. Zigzag Conversion Z 字形变换

字形变换，ZigZag，题解，Leetcode, 力扣，Python, C++, Java

题目地址：https://leetcode.com/problems/zigzag-conversion/description/

题目描述

Thestring "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

Andthen 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 s, int numRows); Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"

Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

题目大意

把一个字符串按照锯齿型的方式去排列，然后按照行进行拼接到一起。输出这样得到的结果。

解题方法

公式

明眼人一看就知道，这个肯定是有公式的。我自己推导的公式和JustDoITopen in new window的一样：

1、第一行和最后一行下标间隔都是interval=n*2-2=8;；
2、中间行的间隔是周期性的,第i行的间隔是:interval–2*i,2*i,interval–2*i,2*i,interval–2*i,2*i,…；

Python 代码如下：

class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        if numRows == 1: return s
        ans = ""
        interval = 2 * (numRows - 1)
        for i in range(0, len(s), interval):
            ans += s[i]
        for row in range(1, numRows - 1):
            inter = 2 * row
            i = row
            while i < len(s):
                ans += s[i]
                inter = interval - inter
                i += inter
        print(ans)
        for i in range(numRows - 1, len(s), interval):
            ans += s[i]
        return ans

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

DDKK.COM 弟弟快看-教程，程序员编程资料站，版权归原作者所有

本文经作者：负雪明烛授权发布，任何组织或个人未经作者授权不得转发