819. Most Common Word 统计高频词(暂未被禁止)
[抄题]:
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.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 是反斜杠不是正斜杠
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
去标点、去空格都用正则表达式
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
- Arrays.asList()返回的是List,而且是一个定长的List。把string数组转成普通数组,才能存到hashset中。
public static void main(String[] args){
2 int[] a1 = new int[]{1,2,3};
3 String[] a2 = new String[]{"a","b","c"};
4
5 System.out.println(Arrays.asList(a1));
6 System.out.println(Arrays.asList(a2));
7 }
打印结果如下:
[[I@dc8569]
[a, b, c]
删除标点、划分空格:反斜杠
String[] words = p.replaceAll("\\pP" , "").toLowerCase().split("\\s+");
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
class Solution {
public String mostCommonWord(String paragraph, String[] banned) {
//ini
Set<String> set = new HashSet<>(Arrays.asList(banned));
Map<String, Integer> map = new HashMap<>();
String res = "";
int max = Integer.MIN_VALUE; //store in HM
String[] words = paragraph.replaceAll("\\pP", "").toLowerCase().split("\\s+");
for (String w : words) {
if (!set.contains(w)) {
map.put(w, map.getOrDefault(w, 0) + 1);
if (map.get(w) > max) {
res = w;
max = map.get(w);
}
}
} //return
return res;
}
}
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