[leetcode]LeetCode刷题的日子--ZigZag Conversion

今天多刷几道题，面试信心就能足一些

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 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

题目翻译：

将一个字符串按照Z形，重新排列输出，具体格式见上文

解题思路：

根据num_row参数，直接构建一个二维数组，然后遍历字符串，每读一个字符，则移动一个单位，从上往下写入时，step为1，由下往上写入时，step为-1，最后顺序获取字符串即可

代码：

fn convert(s: String, num_rows: i32) -> String {
    if num_rows == 1 || num_rows >= s.len() as i32 {
        return s;
    }

    let mut ret: Vec<Vec<char>> = vec![vec![]; num_rows as usize];

    let mut row = 0;
    let mut step: i32 = 1;

    for c in s.chars() {
        ret[row].push(c);
        if row == 0 {
            step = 1;
        }

        if row == num_rows as usize - 1 {
            step = -1;
        }

        row = (row as i32 + step) as usize;
    }

    let mut str_ret = String::new();
    for i in (0..num_rows) {
        for j in &ret[i as usize] {
            str_ret.push(*j);
        }
    }

    return str_ret;
}

#[test]
fn test_convert() {
    let s = String::from("PAYPALISHIRING");
    let row = 3;
    let ret = convert(s, row);

    assert_eq!(ret, "PAHNAPLSIIGYIR");
}