Ai
1 Star 0 Fork 0

徐长贺/Leetcode

加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
文件
克隆/下载
_819.java 2.25 KB
一键复制 编辑 原始数据 按行查看 历史
Steve Sun 提交于 2018-04-16 07:33 +08:00 . add 819
package com.fishercoder.solutions;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
/**
* 819. Most Common Word
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.
1 <= 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 !?',;.
Different words in paragraph are always separated by a space.
There are no hyphens or hyphenated words.
Words only consist of letters, never apostrophes or other punctuation symbols.
*/
public class _819 {
public static class Solution1 {
public String mostCommonWord(String paragraph, String[] banned) {
Set<String> bannedSet = new HashSet(Arrays.asList(banned));
String[] words = paragraph.replaceAll("[^a-zA-Z ]", "").toLowerCase().split("\\s+");
Map<String, Integer> map = new HashMap<>();
Arrays.stream(words)
.filter(word -> !bannedSet.contains(word))
.forEach(word -> map.put(word, map.getOrDefault(word, 0) + 1));
String result = "";
int freq = 0;
for (String key : map.keySet()) {
if (map.get(key) > freq) {
result = key;
freq = map.get(key);
}
}
return result;
}
}
}
Loading...
马建仓 AI 助手
尝试更多
代码解读
代码找茬
代码优化
1
https://gitee.com/isulong/Leetcode.git
git@gitee.com:isulong/Leetcode.git
isulong
Leetcode
Leetcode
master

搜索帮助