原文链接: https://leetcode-cn.com/problems/expressive-words

英文原文

Sometimes people repeat letters to represent extra feeling. For example:

• "hello" -> "heeellooo"
• "hi" -> "hiiii"

In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.

• For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has a size less than three. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If s = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = s.

Return the number of query strings that are stretchy.

 

Example 1:

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Example 2:

Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3

 

Constraints:

• 1 <= s.length, words.length <= 100
• 1 <= words[i].length <= 100
• s and words[i] consist of lowercase letters.

中文题目

有时候人们会用重复写一些字母来表示额外的感受，比如 "hello" -> "heeellooo", "hi" -> "hiii"。我们将相邻字母都相同的一串字符定义为相同字母组，例如："h", "eee", "ll", "ooo"。

对于一个给定的字符串 S ，如果另一个单词能够通过将一些字母组扩张从而使其和 S 相同，我们将这个单词定义为可扩张的（stretchy）。扩张操作定义如下：选择一个字母组（包含字母 c ），然后往其中添加相同的字母 c 使其长度达到 3 或以上。

例如，以 "hello" 为例，我们可以对字母组 "o" 扩张得到 "hellooo"，但是无法以同样的方法得到 "helloo" 因为字母组 "oo" 长度小于 3。此外，我们可以进行另一种扩张 "ll" -> "lllll" 以获得 "helllllooo"。如果 S = "helllllooo"，那么查询词 "hello" 是可扩张的，因为可以对它执行这两种扩张操作使得 query = "hello" -> "hellooo" -> "helllllooo" = S。

输入一组查询单词，输出其中可扩张的单词数量。

 

示例：

输入： 
S = "heeellooo"
words = ["hello", "hi", "helo"]
输出：1
解释：
我们能通过扩张 "hello" 的 "e" 和 "o" 来得到 "heeellooo"。
我们不能通过扩张 "helo" 来得到 "heeellooo" 因为 "ll" 的长度小于 3 。

 

提示：

• 0 <= len(S) <= 100。
• 0 <= len(words) <= 100。
• 0 <= len(words[i]) <= 100。
• S 和所有在 words 中的单词都只由小写字母组成。

通过代码

官方题解

比较相同字母组的长度：

我们首先将 S 拆分成若干组相同的字母，并存储每组字母的长度。例如当 S 为 abbcccddddaaaaa 时，可以得到 5 组字母，它们分别为 abcda，长度为 [1, 2, 3, 4, 5]。

对于 words 中的每个单词 word，如果它可以扩张得到 S，那么它必须和 S 有相同的字母组。对于每一组字母，假设 S 中有 c1 个，word 中有 c2 个，那么会有下面几种情况：

• 如果 c1 < c2，那么 word 不能扩张得到 S；

• 如果 c1 >= 3，那么只要添加 c1 - c2 个字母即可；

• 如果 c1 < 3，由于在扩张时至少需要添加到 3 个字母，所以此时不能添加字母，必须有 c1 == c2。

如果 word 的包含的字母组中的每个字母都满足上述情况，那么 word 可以扩张得到 S。

[sol1]
class Solution {
    public int expressiveWords(String S, String[] words) {
        RLE R = new RLE(S);
        int ans = 0;

        search: for (String word: words) {
            RLE R2 = new RLE(word);
            if (!R.key.equals(R2.key)) continue;
            for (int i = 0; i < R.counts.size(); ++i) {
                int c1 = R.counts.get(i);
                int c2 = R2.counts.get(i);
                if (c1 < 3 && c1 != c2 || c1 < c2)
                    continue search;
            }
            ans++;
        }
        return ans;
    }
}

class RLE {
    String key;
    List<Integer> counts;

    public RLE(String S) {
        StringBuilder sb = new StringBuilder();
        counts = new ArrayList();

        char[] ca = S.toCharArray();
        int N = ca.length;
        int prev = -1;
        for (int i = 0; i < N; ++i) {
            if (i == N-1 || ca[i] != ca[i+1]) {
                sb.append(ca[i]);
                counts.add(i - prev);
                prev = i;
            }
        }

        key = sb.toString();
    }
}

[sol1]
class Solution(object):
    def expressiveWords(self, S, words):
        def RLE(S):
            return zip(*[(k, len(list(grp)))
                         for k, grp in itertools.groupby(S)])

        R, count = RLE(S)
        ans = 0
        for word in words:
            R2, count2 = RLE(word)
            if R2 != R: continue
            ans += all(c1 >= max(c2, 3) or c1 == c2
                       for c1, c2 in zip(count, count2))

        return ans

复杂度分析

• 时间复杂度：$O(N + \sum k_i)$，其中 $N$ 是字符串 S 的长度，$\sum k_i$ 是数组 words 中所有单词的长度之和。

• 空间复杂度：$O(N + K)$，其中 K 是数组 word 中最长的单词的长度。

统计信息

通过次数	提交次数	AC比率
5121	12190	42.0%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言