返回介绍

solution / 1300-1399 / 1324.Print Words Vertically / README_EN

发布于 2024-06-17 01:03:20 字数 4165 浏览 0 评论 0 收藏 0

1324. Print Words Vertically

中文文档

Description

Given a string s. Return all the words vertically in the same order in which they appear in s.
Words are returned as a list of strings, complete with spaces when is necessary. (Trailing spaces are not allowed).
Each word would be put on only one column and that in one column there will be only one word.

 

Example 1:


Input: s = "HOW ARE YOU"

Output: ["HAY","ORO","WEU"]

Explanation: Each word is printed vertically. 

 "HAY"

 "ORO"

 "WEU"

Example 2:


Input: s = "TO BE OR NOT TO BE"

Output: ["TBONTB","OEROOE","   T"]

Explanation: Trailing spaces is not allowed. 

"TBONTB"

"OEROOE"

"   T"

Example 3:


Input: s = "CONTEST IS COMING"

Output: ["CIC","OSO","N M","T I","E N","S G","T"]

 

Constraints:

  • 1 <= s.length <= 200
  • s contains only upper case English letters.
  • It's guaranteed that there is only one space between 2 words.

Solutions

Solution 1

class Solution:
  def printVertically(self, s: str) -> List[str]:
    words = s.split()
    n = max(len(w) for w in words)
    ans = []
    for j in range(n):
      t = [w[j] if j < len(w) else ' ' for w in words]
      while t[-1] == ' ':
        t.pop()
      ans.append(''.join(t))
    return ans
class Solution {
  public List<String> printVertically(String s) {
    String[] words = s.split(" ");
    int n = 0;
    for (var w : words) {
      n = Math.max(n, w.length());
    }
    List<String> ans = new ArrayList<>();
    for (int j = 0; j < n; ++j) {
      StringBuilder t = new StringBuilder();
      for (var w : words) {
        t.append(j < w.length() ? w.charAt(j) : ' ');
      }
      while (t.length() > 0 && t.charAt(t.length() - 1) == ' ') {
        t.deleteCharAt(t.length() - 1);
      }
      ans.add(t.toString());
    }
    return ans;
  }
}
class Solution {
public:
  vector<string> printVertically(string s) {
    stringstream ss(s);
    vector<string> words;
    string word;
    int n = 0;
    while (ss >> word) {
      words.emplace_back(word);
      n = max(n, (int) word.size());
    }
    vector<string> ans;
    for (int j = 0; j < n; ++j) {
      string t;
      for (auto& w : words) {
        t += j < w.size() ? w[j] : ' ';
      }
      while (t.size() && t.back() == ' ') {
        t.pop_back();
      }
      ans.emplace_back(t);
    }
    return ans;
  }
};
func printVertically(s string) (ans []string) {
  words := strings.Split(s, " ")
  n := 0
  for _, w := range words {
    n = max(n, len(w))
  }
  for j := 0; j < n; j++ {
    t := []byte{}
    for _, w := range words {
      if j < len(w) {
        t = append(t, w[j])
      } else {
        t = append(t, ' ')
      }
    }
    for len(t) > 0 && t[len(t)-1] == ' ' {
      t = t[:len(t)-1]
    }
    ans = append(ans, string(t))
  }
  return
}

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。
列表为空,暂无数据
    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
    原文