words

LeetCode 1160. Find Words That Can Be Formed by Characters拼写单词【Easy】【Python】【字符串】 Problem LeetCode You are given an array of strings words and a string chars . A string is good if it can be formed by characters from chars (each character can only be used once). Return the sum of lengths of all good strings in words . Example 1: Input: words = ["cat","bt","hat","tree"], chars = "atach" Output: 6 Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6. Example 2: Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr" Output: 10 Explanation: The

//例1 <script> window.color = 'red'; document.color = 'yellow'; var s1 = {color: 'blue' }; function changeColor(){ console.log(this.color); } changeColor.call(); //red (默认传递参数) changeColor.call(window); //red changeColor.call(document); //yellow changeColor.call(this); //red changeColor.call(s1); //blue </script> //例2 var Pet = { words : '...', speak : function (say) { console.log(say + ''+ this.words) } } Pet.speak('Speak'); // 结果：Speak... var Dog = { words:'Wang' } //将this的指向改变成了Dog Pet.speak.call(Dog, 'Speak'); //结果： SpeakWang //例1 <script> window.number = 'one'; document.number = 'two'; var

思路1 ：遍历，找到每个词的全排列（刚刚做了比较熟悉），然后检查在列表中则删除，继续遍历。–太复杂了。 思路2 ：对字符串进行排序，可检查是否具有相同的字母。然后使用了字典来存储。但是还是比较耗时。后面再看看有没有更简单的做法。 class Solution : def groupAnagrams ( self , strs : List [ str ] ) - > List [ List [ str ] ] : dic = { } for words in strs : b = "" . join ( sorted ( words ) ) if b not in dic : dic [ b ] = [ words ] else : dic [ b ] . append ( words ) return list ( dic . values ( ) ) 来源： CSDN 作者： 三岁与十八 链接： https://blog.csdn.net/qq_27921205/article/details/104159461

Text Analytics-DataCamp-Introduction to Text Analysis in R 1. Wrangling Text 1.1 Text as data (video) 1.2 Airline tweets data Instruction : # Load the tidyverse packages library(tidyverse) # Print twitter_data twitter_data # Print just the complaints in twitter_data twitter_data %>% filter(complaint_label == 'Complaint') 1.3 Grouped summaries Instruction : # Start with the data frame twitter_data %>% # Group the data by whether or not the tweet is a complaint group_by(complaint_label) %>% # Compute the mean, min, and max follower counts summarize( avg_followers = mean(usr_followers_count), min

1 class Solution: 2 def printVertically(self, s: str) -> 'List[str]': 3 words = s.split(' ') 4 matrix = [] 5 maxlen = 0 6 for w in words: 7 maxlen = max(maxlen,len(w)) 8 for word in words: 9 matrix.append(word + ' ' * (maxlen - len(word))) 10 result = [] 11 for j in range(maxlen): 12 temp = '' 13 for i in range(len(words)): 14 temp += matrix[i][j] 15 result.append(temp.rstrip()) 16 return result 先结算出最长的单词的长度，然后组成二维数组，按照先列后行的方式遍历，并且对每一个新组成的单词进行右边去空格处理。 来源： https://www.cnblogs.com/asenyang/p/12215565.html