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 text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
一刷
题解:
根据例子列出n = 3和n = 4的情况，计算出列间距zigSize = 2 * numRows - 2，所以每行元素为 i + n * zigSize以及斜线的计算公式，当i为行数时，斜线上元素为n * zigSize - i。
Time Complexity - O(n)， Space Complexity - O(n)。
public class Solution {
public String convert(String s, int numRows) {
int zigSize = 2*numRows - 2;
if(s == null || s.length() == 0 || zigSize <=0) return s;
StringBuilder res = new StringBuilder();
for(int i=0; i<numRows; i++){ //row
for(int j=0; j<s.length(); j+=zigSize){ //col
if(i + j < s.length()) res.append(s.charAt(i + j));//first row
if(i!=0 && i!= numRows-1 && j + zigSize - i <s.length()){
res.append(s.charAt(j + zigSize - i));
}
}
}
return res.toString();
}
}
