ZigZag Conversion——LeetCode进阶路⑥

原题链接https://leetcode.com/problems/zigzag-conversion/

没开始看题目时，小陌发现这道题似乎备受嫌弃，被n多人踩了，还有点小同情

• 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

• 思路分析：
  题目需求是把输入字符排成锯齿牙子，再按行进行输出
  至于锯齿牙子啥样，没脑补出来的猿兄们看eg：
  Input: s = "123456789", numRows = 2
  1 3 5 7 9
  2 4 6 8
  

Output: "135792468"
Input: s = "123456789", numRows = 3
1     5   9
2  4 6 8
3     7

Output: "159246837"
Input: s = "123456789", numRows = 4
1       7
2    6 8
3 5    9
4

Output: "172683594"
恭喜猿兄，你已经把这道题中“最难”的部分搞定了,如你所见，除了题意相对难理解之外，剩下的简直就是数学的找规律问题
尔后猿兄也发现阿六为啥被踩了叭~
规律分析：(方便起见，n=numRows)

         
首尾两行元素之间的间隔interval1 = 2n-2

两者之间的行中元素的间隔interval = 前一元素的列数 + interval1 - 2 * 当前行数

• 源码附录

  class Solution {
      public String convert(String s, int numRows) {
          if(s == null||numRows<=0){
              return "";
          }
          if(s.length()<2||numRows<2){
              return s;
          }
  
          int interval = 2*numRows - 2;
          int interval1;
          StringBuilder sb = new StringBuilder();
          for(int i=0;i<numRows;i++){
              for(int j=i;j<s.length();j=j+interval){
                  sb.append(s.charAt(j));
                  if(i != 0&&i != numRows-1){
                      interval1 = j + interval - 2*i;
                      if(interval1 < s.length()){
                          sb.append(s.charAt(interval1));
                      }
                  }
              }
          }
          return sb.toString();
      }
  }