返回介绍

solution / 1000-1099 / 1023.Camelcase Matching / README_EN

发布于 2024-06-17 01:03:32 字数 7417 浏览 0 评论 0 收藏 0

1023. Camelcase Matching

中文文档

Description

Given an array of strings queries and a string pattern, return a boolean array answer where answer[i] is true if queries[i] matches pattern, and false otherwise.

A query word queries[i] matches pattern if you can insert lowercase English letters pattern so that it equals the query. You may insert each character at any position and you may not insert any characters.

 

Example 1:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation: "FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".

Example 2:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation: "FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".

Example 3:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation: "FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".

 

Constraints:

  • 1 <= pattern.length, queries.length <= 100
  • 1 <= queries[i].length <= 100
  • queries[i] and pattern consist of English letters.

Solutions

Solution 1: Two Pointers

We can traverse every string in queries and check whether it matches pattern or not. If it matches, we add true to the answer array, otherwise we add false.

Next, we implement a function $check(s, t)$ to check whether the string $s$ matches the string $t$.

We can use two pointers $i$ and $j$ to traverse the two strings. If the characters pointed to by $i$ and $j$ are not the same and $s[i]$ is a lowercase letter, then we move the pointer $i$ to the next position.

If the pointer $i$ has reached the end of the string $s$ or the characters pointed to by $i$ and $j$ are not the same, we return false. Otherwise, we move both pointers $i$ and $j$ to the next position. When the pointer $j$ reaches the end of the string $t$, we need to check if the remaining characters in the string $s$ are all lowercase letters. If so, we return true, otherwise we return false.

Time complexity $(n \times m)$, where $n$ and $m$ are the length of the array queries and the string pattern respectively.

class Solution:
  def camelMatch(self, queries: List[str], pattern: str) -> List[bool]:
    def check(s, t):
      m, n = len(s), len(t)
      i = j = 0
      while j < n:
        while i < m and s[i] != t[j] and s[i].islower():
          i += 1
        if i == m or s[i] != t[j]:
          return False
        i, j = i + 1, j + 1
      while i < m and s[i].islower():
        i += 1
      return i == m

    return [check(q, pattern) for q in queries]
class Solution {
  public List<Boolean> camelMatch(String[] queries, String pattern) {
    List<Boolean> ans = new ArrayList<>();
    for (var q : queries) {
      ans.add(check(q, pattern));
    }
    return ans;
  }

  private boolean check(String s, String t) {
    int m = s.length(), n = t.length();
    int i = 0, j = 0;
    for (; j < n; ++i, ++j) {
      while (i < m && s.charAt(i) != t.charAt(j) && Character.isLowerCase(s.charAt(i))) {
        ++i;
      }
      if (i == m || s.charAt(i) != t.charAt(j)) {
        return false;
      }
    }
    while (i < m && Character.isLowerCase(s.charAt(i))) {
      ++i;
    }
    return i == m;
  }
}
class Solution {
public:
  vector<bool> camelMatch(vector<string>& queries, string pattern) {
    vector<bool> ans;
    auto check = [](string& s, string& t) {
      int m = s.size(), n = t.size();
      int i = 0, j = 0;
      for (; j < n; ++i, ++j) {
        while (i < m && s[i] != t[j] && islower(s[i])) {
          ++i;
        }
        if (i == m || s[i] != t[j]) {
          return false;
        }
      }
      while (i < m && islower(s[i])) {
        ++i;
      }
      return i == m;
    };
    for (auto& q : queries) {
      ans.push_back(check(q, pattern));
    }
    return ans;
  }
};
func camelMatch(queries []string, pattern string) (ans []bool) {
  check := func(s, t string) bool {
    m, n := len(s), len(t)
    i, j := 0, 0
    for ; j < n; i, j = i+1, j+1 {
      for i < m && s[i] != t[j] && (s[i] >= 'a' && s[i] <= 'z') {
        i++
      }
      if i == m || s[i] != t[j] {
        return false
      }
    }
    for i < m && s[i] >= 'a' && s[i] <= 'z' {
      i++
    }
    return i == m
  }
  for _, q := range queries {
    ans = append(ans, check(q, pattern))
  }
  return
}
function camelMatch(queries: string[], pattern: string): boolean[] {
  const check = (s: string, t: string) => {
    const m = s.length;
    const n = t.length;
    let i = 0;
    let j = 0;
    for (; j < n; ++i, ++j) {
      while (i < m && s[i] !== t[j] && s[i].codePointAt(0) >= 97) {
        ++i;
      }
      if (i === m || s[i] !== t[j]) {
        return false;
      }
    }
    while (i < m && s[i].codePointAt(0) >= 97) {
      ++i;
    }
    return i == m;
  };
  const ans: boolean[] = [];
  for (const q of queries) {
    ans.push(check(q, pattern));
  }
  return ans;
}

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