length >= stringArr[1].length) { // 如果第一个元素比第二个长 // 把第二个元素删掉 stringArr.splice(1, 1); // 由于 findLongestWord 接收字符串为参数,因此需要用空格 join return findLongestWord(stringArr.join(" ")); } if (stringArr[0].length < stringArr[1].length) { // 如果第...
String.split() String.length 第一种想法就是,先定一个小变量,来他一个0;然后把每个单词的长度与它比较,把长度大的赋值给该变量,最后返回该变量; functionfindLongestWord(str) {vararray=str.split(' ');//分割句子为单词,保存在数组array里varresult=0;//先定一个小目标for(vari=1;i<array.length;i+...
将这个长度值返回*/functionfindLongestWord(str) {//把字符串 str 转为数组 myarrvarmyarr=str.split(" ");//定义longest方便调用varlongest=0;//遍历数组myarr,myarr.length=6for(vari=0;i<myarr.length;i++){//遍历新数组myarr并将数组中最大的值赋值给longestif(myarr[i].length>longest){ longe...
string = "Hello, world! world is beautiful." positions = [] while True: (tab)pos = string.find("world") (tab)if pos == -1: (2tab)break (tab)positions.append(pos) print(positions) # 输出：[7, 18]判断子字符串是否存在于另一个字符串中 使用find函数返回的结果判断子字符...
capitalise the first letter of each word in a string in SQL Server. Capturing the results from exec command Carriage Return...Line Feed...CHAR(10) and CHAR(13) Help CASE Expression in conjunction with LEN(gln.GLNumber) Case expressions may only be nested to level 10. CASE in JOIN CO...
In cell A2 I have: Field3,Field8,Field9,Field10 In cell B2 I would like to get: 2 (the position of the 2 word Field8 in string) I tried using InStr in UDF In cell B2 I am using below function: FindWordPosition(A2,"Field8") result of function in cell B2= 9 instead of desir...
particular video. The TAG terms under a video goes as a query to this SQL statement to find related videos. Following SQL statement returns only the same video (single video) that has the exact tags. How do I change this to find all videos that has any word of the TAG string. Thanks...
When it is required to find the length of the last word in a string, a method is defined that removes the extra empty spaces in a string, and iterates through the string. It iterates until the last word has been found. Then, its length is found and returned as output. Example Below...
If we are interested in finding the duplicate words along with their count of occureneces in theString, we can use theCollections.frequency(list, item)API that counts the number of times aitemappears in the specifiedlist. Map<String,Integer>dupWordsMapWithCount=newHashMap<>();for(Stringword:...
Hi everbody, I am trying to find specific words from a list of items. I want to know if the word is in the strings given, and if it is, bring to...