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 <= 1000s consists of English letters (lower-case and upper-case), ',' and '.'.1 <= numRows <= 1000struct Solution;
impl Solution {
fn convert(s: String, num_rows: i32) -> String {
let mut res = "".to_string();
let n = num_rows as usize;
if n == 1 {
return s;
}
let m = s.len();
let mut v: Vec<String> = vec!["".to_string(); n];
let mut row = 0;
let mut direction = true;
for j in 0..m {
v[row] += &s[j..=j];
if row == 0 {
direction = true;
row += 1;
} else if row == n - 1 {
direction = false;
row -= 1;
} else {
if direction {
row += 1;
} else {
row -= 1;
}
}
}
for t in v {
res += &t;
}
res
}
}
#[test]
fn test() {
let s = "PAYPALISHIRING".to_string();
let num_rows = 3;
let res = "PAHNAPLSIIGYIR".to_string();
assert_eq!(Solution::convert(s, num_rows), res);
let s = "PAYPALISHIRING".to_string();
let num_rows = 4;
let res = "PINALSIGYAHRPI".to_string();
assert_eq!(Solution::convert(s, num_rows), res);
let s = "AB".to_string();
let num_rows = 1;
let res = "AB".to_string();
assert_eq!(Solution::convert(s, num_rows), res);
}