LeetCode——ZigZag Conversion

本文主要是介绍LeetCode——ZigZag Conversion，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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

题目意思是将字符按Z字型排列 即将0123456789排成

n=3
0		4		8
1	3	5	7	9
2		6
输出0481357926
n=4
0			6
1		5	7
2	4		8
3			9
输出0615724839

相邻两列中多了n-2个数字，因此每一行中的每个数字相邻2n-2，即当n=4时，0和6之间相差24-2.除第一行和最后一行外中间都有数字，中间数字的位置与它前一个数字相差2n-2-2i，i为行数（行数从零开始算）即n=4时，1和5之间相差24-2-21=4。知道它们的位置关系后便可以按顺序添加字符到新字符串中

public String convert(String s, int numRows) {if(numRows<=1)return s;StringBuffer res=new StringBuffer();int size=2*numRows-2;for(int i=0;i<numRows;i++){for(int j=i;j<s.length();j+=size){res.append(s.charAt(j));int tmp=j+size-2*i;if(i!=0&&i!=numRows-1&&tmp<s.length())res.append(s.charAt(tmp));}}return res.toString();}

Runtime: 19 ms, faster than 98.79% of Java online submissions for ZigZag Conversion.
Memory Usage: 39.5 MB, less than 82.45% of Java online submissions for ZigZag Conversion.

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