摘要:并看不懂我不會(huì)到處亂說(shuō)。好吧我努力拆解一下這個(gè)正則是什么意思匹配字符串的開(kāi)始重復(fù)一次或更多次匹配除了以外的任意字符原文
problem:
You will create a program that will find the missing letter from a string and return it. If there is no missing letter, the program should return undefined. There is currently no test case for the string missing more than one letter, but if there was one, recursion would be used. Also, the letters are always provided in order so there is no need to sort them.
fearNotLetter("abce") should return "d".
fearNotLetter("abcdefghjklmno") should return "i".
fearNotLetter("bcd") should return undefined.
fearNotLetter("yz") should return undefined.
var arr = []; var all; for (var i=0; i發(fā)現(xiàn)和上次一樣,我在進(jìn)入循環(huán)什么時(shí)候return出來(lái)這里總是做錯(cuò),或者想的復(fù)雜... 總結(jié)一下:return之后會(huì)直接結(jié)束function;如果for循環(huán)一變不出return直接外面來(lái)return undefined也是可以的。 basic solution: function fearNotLetter(str) { for(var i = 0; i < str.length; i++) { var code = str.charCodeAt(i); if (code !== str.charCodeAt(0) + i) { return String.fromCharCode(code - 1); } } return undefined; }Intermediate Code Solution:function fearNotLetter(str) { var compare = str.charCodeAt(0), missing; str.split("").map(function(letter,index) { if (str.charCodeAt(index) == compare) { ++compare; } else { missing = String.fromCharCode(compare); } }); return missing; }Advanced Code Solution:function fearNotLetter(str) { var allChars = ""; var notChars = new RegExp("[^"+str+"]","g"); for (var i = 0; allChars[allChars.length-1] !== str[str.length-1] ; i++) allChars += String.fromCharCode(str[0].charCodeAt(0) + i); return allChars.match(notChars) ? allChars.match(notChars).join("") : undefined; }并看不懂我不會(huì)到處亂說(shuō)。好吧我努力拆解一下這個(gè)正則是什么意思...
^ 匹配字符串的開(kāi)始 + 重復(fù)一次或更多次 [^x] 匹配除了x以外的任意字符a regular expression notChars which selects everything except str
?: Conditional(ternary)Operatorhttps://developer.mozilla.org...
Syntax:condition ? expr1 : expr2Parameters:
condition (or conditions) An expression that evaluates to true or false.
expr1, expr2 Expressions with values of any type.原文:https://forum.freecodecamp.or...
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摘要:更多的小算法練習(xí),可以查看我的文章。規(guī)則使用語(yǔ)言,使用函數(shù)獲取傳遞的參數(shù)并使用以下算法對(duì)其進(jìn)行修改。將字符串中的每個(gè)字母替換為字母表后面的字母即變?yōu)椋優(yōu)?。然后將這個(gè)新字符串,,,,中的每個(gè)元音大寫,并最終返回此修改后的字符串。 雖然都是很簡(jiǎn)單的算法,每個(gè)都只需5分鐘左右,但寫起來(lái)總會(huì)遇到不同的小問(wèn)題,希望大家能跟我一起每天進(jìn)步一點(diǎn)點(diǎn)。更多的小算法練習(xí),可以查看我的文章。 規(guī)則 Usi...
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