zoukankan      html  css  js  c++  java
  • [leetcode]819. Most Common Word 出现频率最高的non-banned单词

    Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words.  It is guaranteed there is at least one word that isn't banned, and that the answer is unique.

    Words in the list of banned words are given in lowercase, and free of punctuation.  Words in the paragraph are not case sensitive.  The answer is in lowercase.

    Example:

    Input: 
    paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
    banned = ["hit"]
    Output: "ball"
    Explanation: 
    "hit" occurs 3 times, but it is a banned word.
    "ball" occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph. 
    Note that words in the paragraph are not case sensitive,
    that punctuation is ignored (even if adjacent to words, such as "ball,"), 
    and that "hit" isn't the answer even though it occurs more because it is banned.

    Note:

    • 1 <= paragraph.length <= 1000.
    • 0 <= banned.length <= 100.
    • 1 <= banned[i].length <= 10.
    • The answer is unique, and written in lowercase (even if its occurrences in paragraph may have uppercase symbols, and even if it is a proper noun.)
    • paragraph only consists of letters, spaces, or the punctuation symbols !?',;.
    • There are no hyphens or hyphenated words.
    • Words only consist of letters, never apostrophes or other punctuation symbols.

    题意

    求出现频率最高的non-banned单词

    思路

    HashMap

    代码

    class Solution {  
        public String mostCommonWord(String paragraph, String[] banned) {
            String[] words = paragraph.replaceAll("\pP" , " ").toLowerCase().split("\s+");
            Map<String, Integer> map = new HashMap<>();
            for(String word : words){
                if(map.containsKey(word)){
                    map.put(word, map.get(word) + 1);
                }else{
                    map.put(word, 1);
                }
            }      
            for(String bword : banned) {
                if(map.containsKey(bword)){
                    map.remove(bword); 
                }
            }
            String res = "";
            for(String word : map.keySet())
                if(res == "" || map.get(word) > map.get(res))
                    res = word;
            return res;
        }
    }
  • 相关阅读:
    Google Guava官方教程(中文版)
    maven POM.xml 标签详解
    Linux内存文件系统tmpfs(/dev/shm)详细介绍
    linux 如何查看目录的剩余空间大小?
    CSS position绝对定位absolute relative
    java 笔记
    js笔记
    AFNetworking监控网络状态以及下载的进度条显示
    HTTPS网络请求时plist文件的设置
    KVO监测tableView的滑动
  • 原文地址:https://www.cnblogs.com/liuliu5151/p/11124121.html
Copyright © 2011-2022 走看看