Title description:
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:
IInput: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:
P _ _ I _ _ N
A _ L S _ I G
Y A _ H R _ _
P _ _ I _ _ _
Example 3:
Input: s = “A”, numRows = 1
Output: “A”
Constraints:
1 <= s.length <= 1000
s consists of English letters (lower-case and upper-case), ‘,’ and ‘.’.
1 <= numRows <= 1000
Time complexity: O(n)
to find the law:
1. For the first row and the last row, it is an arithmetic sequence with a tolerance of 2(n−1), and the first term is 0 and n−1;
2. For other rows (0 <i<n−1), is the alternate arrangement of two arithmetic series with tolerance of 2(n−1), the first term is i and 2n−i−2 respectively;
class Solution {
public String convert(String s, int numRows) {
if(numRows == 1) return s;
StringBuilder res = new StringBuilder();
int len = s.length();
int n = numRows;
for(int i = 0; i < n; i++){
if(i == 0 || i == n-1){
for(int j = i; j < len; j+= 2*n-2){
res.append(s.charAt(j));
}
}else{
for(int j = i, k = 2*n -2 - i; j < len || k < len; j+= 2*n-2, k+=2*n-2 ){
if(j < len) res.append(s.charAt(j));
if(k < len) res.append(s.charAt(k));
}
}
}
return res.toString();
}
}