Question

6. Z 字形变换

English Version

题目描述

将一个给定字符串 s 根据给定的行数 numRows ，以从上往下、从左到右进行 Z 字形排列。

比如输入字符串为 "PAYPALISHIRING" 行数为 3 时，排列如下：

P   A   H   N
A P L S I I G
Y   I   R

之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如："PAHNAPLSIIGYIR"。

请你实现这个将字符串进行指定行数变换的函数：

string convert(string s, int numRows);

 

示例 1：

输入：s = "PAYPALISHIRING", numRows = 3
输出："PAHNAPLSIIGYIR"

示例 2：

输入：s = "PAYPALISHIRING", numRows = 4
输出："PINALSIGYAHRPI"
解释：
P   I  N
A   L S  I G
Y A   H R
P   I

示例 3：

输入：s = "A", numRows = 1
输出："A"

 

提示：

• 1 <= s.length <= 1000
• s 由英文字母（小写和大写）、',' 和 '.' 组成
• 1 <= numRows <= 1000

解法

方法一：模拟

我们用一个二维数组 $g$ 来模拟 $Z$ 字形排列的过程，其中 $g[i][j]$ 表示第 $i$ 行第 $j$ 列的字符。初始时 $i=0$，另外我们定义一个方向变量 $k$，初始时 $k=-1$，表示向上走。

我们从左到右遍历字符串 $s$，每次遍历到一个字符 $c$，将其追加到 $g[i]$ 中，如果此时 $i=0$ 或者 $i=numRows-1$，说明当前字符位于 $Z$ 字形排列的拐点，我们将 $k$ 的值反转，即 $k=-k$。接下来，我们将 $i$ 的值更新为 $i+k$，即向上或向下移动一行。继续遍历下一个字符，直到遍历完字符串 $s$，我们返回 $g$ 中所有行拼接后的字符串即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。

class Solution:
  def convert(self, s: str, numRows: int) -> str:
    if numRows == 1:
      return s
    g = [[] for _ in range(numRows)]
    i, k = 0, -1
    for c in s:
      g[i].append(c)
      if i == 0 or i == numRows - 1:
        k = -k
      i += k
    return ''.join(chain(*g))

class Solution {
  public String convert(String s, int numRows) {
    if (numRows == 1) {
      return s;
    }
    StringBuilder[] g = new StringBuilder[numRows];
    Arrays.setAll(g, k -> new StringBuilder());
    int i = 0, k = -1;
    for (char c : s.toCharArray()) {
      g[i].append(c);
      if (i == 0 || i == numRows - 1) {
        k = -k;
      }
      i += k;
    }
    return String.join("", g);
  }
}

class Solution {
public:
  string convert(string s, int numRows) {
    if (numRows == 1) {
      return s;
    }
    vector<string> g(numRows);
    int i = 0, k = -1;
    for (char c : s) {
      g[i] += c;
      if (i == 0 || i == numRows - 1) {
        k = -k;
      }
      i += k;
    }
    string ans;
    for (auto& t : g) {
      ans += t;
    }
    return ans;
  }
};

func convert(s string, numRows int) string {
  if numRows == 1 {
    return s
  }
  g := make([][]byte, numRows)
  i, k := 0, -1
  for _, c := range s {
    g[i] = append(g[i], byte(c))
    if i == 0 || i == numRows-1 {
      k = -k
    }
    i += k
  }
  return string(bytes.Join(g, nil))
}

function convert(s: string, numRows: number): string {
  if (numRows === 1) {
    return s;
  }
  const g: string[][] = new Array(numRows).fill(0).map(() => []);
  let i = 0;
  let k = -1;
  for (const c of s) {
    g[i].push(c);
    if (i === numRows - 1 || i === 0) {
      k = -k;
    }
    i += k;
  }
  return g.flat().join('');
}

impl Solution {
  pub fn convert(s: String, num_rows: i32) -> String {
    let num_rows = num_rows as usize;
    if num_rows == 1 {
      return s;
    }
    let mut ss = vec![String::new(); num_rows];
    let mut i = 0;
    let mut to_down = true;
    for c in s.chars() {
      ss[i].push(c);
      if to_down {
        i += 1;
      } else {
        i -= 1;
      }
      if i == 0 || i == num_rows - 1 {
        to_down = !to_down;
      }
    }
    let mut res = String::new();
    for i in 0..num_rows {
      res += &ss[i];
    }
    res
  }
}

/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function (s, numRows) {
  if (numRows === 1) {
    return s;
  }
  const g = new Array(numRows).fill(_).map(() => []);
  let i = 0;
  let k = -1;
  for (const c of s) {
    g[i].push(c);
    if (i === 0 || i === numRows - 1) {
      k = -k;
    }
    i += k;
  }
  return g.flat().join('');
};

public class Solution {
  public string Convert(string s, int numRows) {
    if (numRows == 1) {
      return s;
    }
    int n = s.Length;
    StringBuilder[] g = new StringBuilder[numRows];
    for (int j = 0; j < numRows; ++j) {
      g[j] = new StringBuilder();
    }
    int i = 0, k = -1;
    foreach (char c in s.ToCharArray()) {
      g[i].Append(c);
      if (i == 0 || i == numRows - 1) {
        k = -k;
      }
      i += k;
    }
    StringBuilder ans = new StringBuilder();
    foreach (StringBuilder t in g) {
      ans.Append(t);
    }
    return ans.ToString();
  }
}

方法二

class Solution:
  def convert(self, s: str, numRows: int) -> str:
    if numRows == 1:
      return s
    group = 2 * numRows - 2
    ans = []
    for i in range(1, numRows + 1):
      interval = group if i == numRows else 2 * numRows - 2 * i
      idx = i - 1
      while idx < len(s):
        ans.append(s[idx])
        idx += interval
        interval = group - interval
        if interval == 0:
          interval = group
    return ''.join(ans)

class Solution {
  public String convert(String s, int numRows) {
    if (numRows == 1) {
      return s;
    }
    StringBuilder ans = new StringBuilder();
    int group = 2 * numRows - 2;
    for (int i = 1; i <= numRows; i++) {
      int interval = i == numRows ? group : 2 * numRows - 2 * i;
      int idx = i - 1;
      while (idx < s.length()) {
        ans.append(s.charAt(idx));
        idx += interval;
        interval = group - interval;
        if (interval == 0) {
          interval = group;
        }
      }
    }
    return ans.toString();
  }
}

class Solution {
public:
  string convert(string s, int numRows) {
    if (numRows == 1) return s;
    string ans;
    int group = 2 * numRows - 2;
    for (int i = 1; i <= numRows; ++i) {
      int interval = i == numRows ? group : 2 * numRows - 2 * i;
      int idx = i - 1;
      while (idx < s.length()) {
        ans.push_back(s[idx]);
        idx += interval;
        interval = group - interval;
        if (interval == 0) interval = group;
      }
    }
    return ans;
  }
};

func convert(s string, numRows int) string {
  if numRows == 1 {
    return s
  }
  n := len(s)
  ans := make([]byte, n)
  step := 2*numRows - 2
  count := 0
  for i := 0; i < numRows; i++ {
    for j := 0; j+i < n; j += step {
      ans[count] = s[i+j]
      count++
      if i != 0 && i != numRows-1 && j+step-i < n {
        ans[count] = s[j+step-i]
        count++
      }
    }
  }
  return string(ans)
}

function convert(s: string, numRows: number): string {
  if (numRows === 1) {
    return s;
  }
  const ss = new Array(numRows).fill('');
  let i = 0;
  let toDown = true;
  for (const c of s) {
    ss[i] += c;
    if (toDown) {
      i++;
    } else {
      i--;
    }
    if (i === 0 || i === numRows - 1) {
      toDown = !toDown;
    }
  }
  return ss.reduce((r, s) => r + s);
}

impl Solution {
  pub fn convert(s: String, num_rows: i32) -> String {
    let num_rows = num_rows as usize;
    let mut rows = vec![String::new(); num_rows];
    let iter = (0..num_rows).chain((1..num_rows - 1).rev()).cycle();
    iter.zip(s.chars()).for_each(|(i, c)| rows[i].push(c));
    rows.into_iter().collect()
  }
}

/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function (s, numRows) {
  if (numRows == 1) return s;
  const arr = new Array(numRows);
  for (let i = 0; i < numRows; i++) arr[i] = [];
  let mi = 0,
    isDown = true;
  for (const c of s) {
    arr[mi].push(c);

    if (mi >= numRows - 1) isDown = false;
    else if (mi <= 0) isDown = true;

    if (isDown) mi++;
    else mi--;
  }
  let ans = [];
  for (const item of arr) {
    ans = ans.concat(item);
  }
  return ans.join('');
};

class Solution {
  /**
   * @param string $s
   * @param int $numRows
   * @return string
   */

  function convert($s, $numRows) {
    if ($numRows == 1 || strlen($s) <= $numRows) {
      return $s;
    }

    $result = '';
    $cycleLength = 2 * $numRows - 2;
    $n = strlen($s);

    for ($i = 0; $i < $numRows; $i++) {
      for ($j = 0; $j + $i < $n; $j += $cycleLength) {
        $result .= $s[$j + $i];

        if ($i != 0 && $i != $numRows - 1 && $j + $cycleLength - $i < $n) {
          $result .= $s[$j + $cycleLength - $i];
        }
      }
    }

    return $result;
  }
}