6. ZigZag Conversion Z 字形变换


作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 公众号:负雪明烛 本文关键词:字形变换,ZigZag,题解,Leetcode, 力扣,Python, C++, Java


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

题目描述

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 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

日期

2018 年 6 月 27 日 ———— 阳光明媚,心情大好,抓紧科研啊