zoukankan      html  css  js  c++  java
  • 748. Shortest Completing Word 最短的匹配单词

     Find the minimum length word from a given dictionary words, which has all the letters from the string licensePlate. Such a word is said to complete the given string licensePlate

    Here, for letters we ignore case. For example, "P" on the licensePlate still matches "p" on the word.

    It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.

    The license plate might have the same letter occurring multiple times. For example, given a licensePlate of "PP", the word "pair" does not complete the licensePlate, but the word "supper" does.

    Example 1:

    Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
    Output: "steps"
    Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
    Note that the answer is not "step", because the letter "s" must occur in the word twice.
    Also note that we ignored case for the purposes of comparing whether a letter exists in the word.
    

    Example 2:

    Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
    Output: "pest"
    Explanation: There are 3 smallest length words that contains the letters "s".
    We return the one that occurred first.
    

    Note:

    1. licensePlate will be a string with length in range [1, 7].
    2. licensePlate will contain digits, spaces, or letters (uppercase or lowercase).
    3. words will have a length in the range [10, 1000].
    4. Every words[i] will consist of lowercase letters, and have length in range [1, 15].

    1. /**
    2. * @param {string} licensePlate
    3. * @param {string[]} words
    4. * @return {string}
    5. */
    6. var shortestCompletingWord = function (licensePlate, words) {
    7. let m = {};
    8. for (let i = 0; i < licensePlate.length; i++) {
    9. let c = licensePlate[i].toLowerCase();
    10. if (!isNaN(c)) continue;
    11. m[c] = m[c] ? ++m[c] : 1;
    12. }
    13. let res = "";
    14. for (let i in words) {
    15. let str = words[i];
    16. if (isMatch(m, str) && (str.length < res.length || res == "")) {
    17. res = str;
    18. }
    19. }
    20. return res;
    21. };
    22. var isMatch = (m, s) => {
    23. let newM = {};
    24. for (let i = 0; i < s.length; i++) {
    25. let c = s[i].toLowerCase();
    26. if (!isNaN(c)) continue;
    27. newM[c] = newM[c] ? ++newM[c] : 1;
    28. }
    29. for (let i in m) {
    30. if (!newM[i] || m[i] > newM[i]) {
    31. return false;
    32. }
    33. }
    34. return true;
    35. }
    36. let licensePlate = "GrC8950";
    37. let words = ["measure", "other", "every", "base", "according", "level", "meeting", "none", "marriage", "rest"];
    38. console.log(shortestCompletingWord(licensePlate, words));






  • 相关阅读:
    转:超级好用的流程图js框架
    流程图插件
    转:介绍几个著名的实用的Java反编译工具,提供下载
    关于 web.config impersonate 帐号模拟
    SQLSERVER 使用 ROLLUP 汇总数据,实现分组统计,总计(合计),小计
    【论文排版工具】——LaTeX的安装及使用(MiKTeX+TexStudio+Windows)
    C语言输入带空格的字符串
    SQL-连接查询:left join,right join,inner join,full join之间的区别
    MySQL与Oracle的隔离级别
    区块链节点运维相关
  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/8083018.html
Copyright © 2011-2022 走看看