Medium
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
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
impl Solution {
pub fn convert(s: String, num_rows: i32) -> String {
let num_rows = num_rows as usize;
if num_rows == 1 || s.len() < num_rows {
return s;
}
let mut result = String::with_capacity(s.len());
let s_bytes = s.as_bytes();
let complete_jump = 2 * (num_rows - 1);
for row in 0..num_rows {
let mut index = row;
let zig_jump = 2 * (num_rows - 1 - row);
let zag_jump = complete_jump - zig_jump;
let mut use_zig_jump = true;
while index < s.len() {
result.push(s_bytes[index] as char);
if row == 0 || row == num_rows - 1 {
index += complete_jump;
} else {
index += if use_zig_jump { zig_jump } else { zag_jump };
use_zig_jump = !use_zig_jump;
}
}
}
result
}
}