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LeetCode 解题报告(362) -- 745. Prefix and Suffix Search

Problem

Design a special dictionary which has some words and allows you to search the words in it by a prefix and a suffix.

Implement the WordFilter class:

  • WordFilter (string [] words) Initializes the object with the words in the dictionary.
  • f (string prefix, string suffix) Returns the index of the word in the dictionary which has the prefix prefix and the suffix suffix. If there is more than one valid index, return the largest of them. If there is no such word in the dictionary, return -1.

Example 1:

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Input
["WordFilter", "f"]
[[["apple"]], ["a", "e"]]
Output
[null, 0]

Explanation
WordFilter wordFilter = new WordFilter (["apple"]);
wordFilter.f ("a", "e"); // return 0, because the word at index 0 has prefix = "a" and suffix = 'e".

Constraints:

  • 1 <= words.length <= 15000
  • 1 <= words [i].length <= 10
  • 1 <= prefix.length, suffix.length <= 10
  • words [i], prefix and suffix consist of lower-case English letters only.
  • At most 15000 calls will be made to the function f.

Analysis

   题目给出一系列单词,然后要求我们实现一个前后缀的搜索功能,当调用搜索方法时,传入前后缀 prefixsuffix,返回符合要求的单词,如果有多个单词都符合要求,则返回下标最大的。

   首先对每个单词都找出它所有的前缀和后缀,然后用一个非字母的符号把它们两两串联起来,这样的话,在查询的时候,给出前缀和后缀就能快速查找到对应的单词。

   由于在有多个单词都匹配的情况下需要返回下标最大的,所以这里我使用了一个 map 去做存储,key 是匹配的字符串,value 是最大的下标。从左往右遍历,这样能够保证匹配字符串相同的情况下,value 是最大的下标。


Solution

   无。


Code

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class WordFilter {
public:
WordFilter (vector<string>& words) {
int size = words.size ();
for (int i = 0; i < size; i++) {
int s = words [i].size ();
for (int j = 0; j <= s; j++) {
for (int k = s; k >= 0; k--) {
m [words [i].substr (0, j) + "#" + words [i].substr (k)] = i;
}
}
}
}

int f(string prefix, string suffix) {
return (m.count (prefix + "#" + suffix)) ? m [prefix + "#" + suffix]: -1;
}
private:
unordered_map<string, int> m;
};

/**
* Your WordFilter object will be instantiated and called as such:
* WordFilter* obj = new WordFilter (words);
* int param_1 = obj->f (prefix,suffix);
*/

Summary

   这道题目也是比较典型的字符串匹配类型题目,也是通过使用非字母字符的形式,构造出类似正则表达式的字符串,然后存放在 map 中用于匹配。这道题目的分享到这里,感谢你的支持!

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