6.ZigZag Conversion

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题目

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   NA P L S I I GY   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".

解题思路

解法有很多，这里采用的方法是这样的：首先算出变换之后每一行的起始地址，然后再一组一组地变换，每组元素的个数为2*numRows－2。时空复杂度均为O(n) 。

代码

public class Solution {public String convert(String s,int numRows){if(numRows<=1 || s.length()==0)return s;char[] ss=new char[s.length()];int[] startIndex=new int[numRows];int len=2*numRows-2;intn=s.length()/len;intm=s.length()%len;startIndex[0]=0;//the start index of each row after convertstartIndex[1]=startIndex[0]+n+(m>=1?1:0);for(int i=2;i<numRows;++i){startIndex[i]=startIndex[i-1]+2*n;if(m>=i)++startIndex[i];if(m>=2*numRows-i)++startIndex[i];}for(int i=0;i*len<s.length();++i)//fill in the answer string group by group(2*numRows-2){ss[startIndex[0]++]=s.charAt(i*len);for(int j=1;j<numRows-1;++j){if(i*len+j<s.length())ss[startIndex[j]++]=s.charAt(i*len+j);if(i*len+2*numRows-2-j<s.length())ss[startIndex[j]++]=s.charAt(i*len+2*numRows-2-j);}if(i*len+numRows-1<s.length())ss[startIndex[numRows-1]++]=s.charAt(i*len+numRows-1);}return String.valueOf(ss);    }｝

运行结果

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