字符串操作大全:面试准备和日常编码所需一文打尽!

图源:unsplash

字符串是一系列字符,由常数或变量构成。它是编程语言中必不可少的数据类型。本文中将重点关注JavaScript字符串操作,但其原理和算法也可应用于其他语言。

参加技术面试时,面试官常常会关注以下内容:

· 编程技术

· 语言能力

· 解题技巧

本文不仅可以让你成功通过技术面试,对日常编码也很有用。代码要点格式中,我们列出了JavaScript字符串的几点重要特性,这是编程技能的基础。其中包括存在了20年的属性和方法,也涵盖ES2021中的特性。如有不清楚之处,可以有针对性地查漏补缺。JavaScript编程语言可以解决许多应用问题。这些算法或其变体,经常出现在真实的面试场景中。

字符串属性和方法

字符串用于表示和操作字符序列。字符串属性和方法有很多。以下是可供参考的代码示例,包括ES2020中的“matchAll”和ES2021中的“replaceAll”。

const str = Today is a nice day! ;
console.log(str.length); // 20
console.log(str[2]); // "d"
console.log(typeof str); // "string"
console.log(typeof str[2]); // "string"
console.log(typeofString(5)); //"string"
console.log(typeofnewString(str)); //"object"
console.log(str.indexOf( is )); // 6
console.log(str.indexOf( today )); // -1
console.log(str.includes( is )); // true
console.log(str.includes( IS )); // false
console.log(str.startsWith( Today )); // true
console.log(str.endsWith( day )); // false
console.log(str.split(   )); // ["Today", "is", "a", "nice","day!"]
console.log(str.split(  )); // ["T", "o", "d", "a","y", " ", "i", "s", " ","a", " ", "n", "i", "c","e", " ", "d", "a", "y","!"]
console.log(str.split( a )); // ["Tod", "y is ", " nice d","y!"]
console.log(str +1+2); // "Today is a nice day!12"
console.log(str + str); // "Today is a nice day!Today is a niceday!"
console.log(str.concat(str)); // "Today is a nice day!Today is a niceday!"
console.log(str.repeat(2)); // "Today is a nice day!Today is a nice day!"
console.log( abc < bcd ); // true
console.log( abc .localeCompare( bcd )); // -1
console.log( a .localeCompare( A )); // -1
console.log( a .localeCompare( A , undefined, { numeric: true })); // -1
console.log( a .localeCompare( A , undefined, { sensitivity:  accent  })); // 0
console.log( a .localeCompare( A , undefined, { sensitivity:  base  })); // 0
console.log( a .localeCompare( A! , undefined, { sensitivity:  base , ignorePunctuation: true })); // 0
console.log( abc .toLocaleUpperCase()); // "ABC"
console.log(str.padStart(25,  * )); // "*****Todayis a nice day!"
console.log(str.padEnd(22,  ! )); // "Today is anice day!!!"
console.log(    middle  .trim().length); // 6
console.log(    middle  .trimStart().length); // 8
console.log(    middle  .trimEnd().length); // 9
console.log(str.slice(6, 8)); // "is"
console.log(str.slice(-4)); // "day!"
console.log(str.substring(6, 8)); // "is"
console.log(str.substring(-4)); // "Today is a nice day!"
console.log( a .charCodeAt()); // 97
console.log(String.fromCharCode(97)); // "a"
console.log(str.search(/[a-c]/)); // 3
console.log(str.match(/[a-c]/g)); // ["a", "a", "c", "a"]
console.log([...str.matchAll(/[a-c]/g)]);
// [Array(1), Array(1), Array(1), Array(1)]
// 0: ["a", index: 3, input: "Today is a nice day!",groups: undefined]
// 1: ["a", index: 9, input: "Today is a nice day!",groups: undefined]
// 2: ["c", index: 13, input: "Today is a niceday!", groups: undefined]
// 3: ["a", index: 17, input: "Today is a niceday!", groups: undefined]
console.log([... test1test2 .matchAll(/t(e)(st(d?))/g)]);
// [Array(4), Array(4)]
// 0: (4) ["test1", "e", "st1","1", index: 0, input: "test1test2", groups: undefined]
// 1: (4) ["test2", "e", "st2","2", index: 5, input: "test1test2", groups: undefined]
console.log(str.replace( a ,  z )); // Todzy is anice day!
console.log(str.replace(/[a-c]/,  z )); // Todzy is anice day!
console.log(str.replace(/[a-c]/g,  z )); // Todzy is znize dzy!
console.log(str.replaceAll( a ,  z )); // Todzy is znice dzy!
console.log(str.replaceAll(/[a-c]/g,  z )); // Todzy is znize dzy!
console.log(str.replaceAll(/[a-c]/,  z )); // TypeError:String.prototype.replaceAll called with a non-global RegExp argument

映射和集合

对于字符串操作,我们需要在某处存储中间值。数组、映射和集合都是需要掌握的常用数据结构,本文主要讨论集合和映射。

集合

Set是存储所有类型的唯一值的对象。以下是供参考的代码示例,一目了然。

const set =newSet( aabbccdefghi );
console.log(set.size); // 9
console.log(set.has( d )); // true
console.log(set.has( k )); // false
console.log(set.add( k )); // {"a", "b", "c", "d","e" "f", "g", "h", "i","k"}
console.log(set.has( k )); // true
console.log(set.delete( d )); // true
console.log(set.has( d )); // false
console.log(set.keys()); // {"a", "b", "c","e" "f", "g", "h", "i","k"}
console.log(set.values()); // {"a", "b", "c","e" "f", "g", "h", "i","k"}
console.log(set.entries()); // {"a" => "a","b" => "b", "c" => "c","e" => "e",
// "f"=> "f", "g" => "g", "h" =>"h"}, "i" => "i", "k" =>"k"}
const set2 =newSet();
set.forEach(item => set2.add(item.toLocaleUpperCase()));
set.clear();
console.log(set); // {}
console.log(set2); //{"A", "B", "C", "E", "F","G", "H", "I", "K"}
console.log(newSet([{ a: 1, b: 2, c: 3 }, { d: 4, e: 5 }, { d: 4, e: 5 }]));
// {{a: 1, b: 2,c: 3}, {d: 4, e: 5}, {d: 4, e: 5}}
const item = { f: 6, g: 7 };
console.log(newSet([{ a: 1, b: 2, c: 3 }, item, item]));
// {{a: 1, b: 2,c: 3}, {f: 6, g: 7}}

映射

映射是保存键值对的对象。任何值都可以用作键或值。映射会记住键的原始插入顺序。以下是供参考的代码示例:

value = ${value}, key = ${key}, map = ${map.size}

应用题

面试中有英语应用题,我们探索了一些经常用于测试的算法。

等值线

等值线图是指所含字母均只出现一次的单词。

· dermatoglyphics (15个字母)

· hydropneumatics (15个字母)

· misconjugatedly (15个字母)

· uncopyrightable (15个字母)

· uncopyrightables (16个字母)

· subdermatoglyphic (17个字母)

如何写一个算法来检测字符串是否是等值线图?有很多方法可以实现。可以把字符串放在集合中,然后自动拆分成字符。由于集合是存储唯一值的对象,如果它是一个等值线图,它的大小应该与字符串长度相同。

/**
* An algorithm to verify whethera given string is an isogram
* @param {string} str The string to be verified
* @return {boolean} Returns whether it is an isogram
*/
functionisIsogram(str) {
if (!str) {
returnfalse;
}
const set =newSet(str);
return set.size=== str.length;
}

以下是验证测试:

console.log(isIsogram(  )); // false
console.log(isIsogram( a )); // true
console.log(isIsogram( misconjugatedly )); // true
console.log(isIsogram( misconjugatledly )); // false

全字母短句

全字母短句是包含字母表中所有26个字母的句子,不分大小写。理想情况下,句子越短越好。以下为全字母短句:

· Waltz, bad nymph, for quick jigs vex. (28个字母)

· Jived fox nymph grabs quick waltz. (28个字母)

· Glib jocks quiz nymph to vex dwarf. (28个字母)

· Sphinx of black quartz, judge my vow. (29个字母)

· How vexingly quick daft zebras jump! (30个字母)

· The five boxing wizards jump quickly. (31个字母)

· Jackdaws love my big sphinx of quartz. (31个字母)

· Pack my box with five dozen liquor jugs. (32个字母)

· The quick brown fox jumps over a lazy dog. (33个字母)

还有很多方法可以验证给定的字符串是否是全字母短句。这一次,我们将每个字母(转换为小写)放入映射中。如果映射大小为26,那么它就是全字母短句。

/**
* An algorithm to verify whethera given string is a pangram
* @param {string} str The string to be verified
* @return {boolean} Returns whether it is a pangram
*/
functionisPangram(str) {
const len = str.length;
if (len <26) {
returnfalse;
}
const map =newMap();
for (let i =0; i < len; i++) {
if (str[i].match(/[a-z]/i)) { // if it is letter a to z, ignoring the case
map.set(str[i].toLocaleLowerCase(), true); // use lower case letter as a key
}
}
return map.size===26;
}

以下是验证测试:

console.log(isPangram(  )); // false
console.log(isPangram( Bawds jog, flick quartz, vex nymphs. )); // true
console.log(isPangram( The quick brown fox jumped over the lazy sleepingdog. )); // true
console.log(isPangram( Roses are red, violets are blue, sugar is sweet,and so are you. )); // false

同构字符串

给定两个字符串s和t,如果可以替换掉s中的字符得到t,那么这两个字符串是同构的。s中的所有字符转换都必须应用到s中相同的字符上,例如,murmur与tartar为同构字符串,如果m被t替换,u被a替换,r被自身替换。以下算法使用数组来存储转换字符,也适用于映射。

/**
* An algorithm to verify whethertwo given strings are isomorphic
* @param {string} s The first string
* @param {string} t The second string
* @return {boolean} Returns whether these two strings are isomorphic
*/
functionareIsomorphic(s, t) {
// strings with different lengths are notisomorphic
if (s.length !== t.length) {
returnfalse;
}
// the conversion array
const convert = [];
for (let i =0; i < s.length; i++) {
// if the conversioncharacter exists
if (convert[s[i]]) {
// apply the conversion and compare
if (t[i] === convert[s[i]]) { // so far so good
continue;
}
returnfalse; // not isomorphic
}
// set the conversion character for future use
convert[s[i]] = t[i];
}
// these two strings are isomorphic since there are no violations
returntrue;
};

以下是验证测试:

console.log(areIsomorphic( atlatl ,  tartar )); // true
console.log(areIsomorphic( atlatlp ,  tartarq )); // true
console.log(areIsomorphic( atlatlpb ,  tartarqc )); // true
console.log(areIsomorphic( atlatlpa ,  tartarqb )); // false

相同字母异构词

相同字母异构词是通过重新排列不同单词的字母而形成的单词,通常使用所有原始字母一次。从一个池中重新排列单词有很多种可能性。例如,cat的相同字母异构词有cat、act、atc、tca、atc和tac。我们可以添加额外的要求,即新单词必须出现在源字符串中。如果源实际上是actually,则结果数组是[“act”]。

/**
* Given a pool to compose ananagram, show all anagrams contained (continuously) in the source
* @param {string} source A source string to draw an anagram from
* @param {string} pool A pool to compose an anagram
* @return {array} Returns an array of anagrams that are contained by the source string
*/
functionshowAnagrams(source, pool) {
// if source is not long enough to hold theanagram
if (source.length< pool.length) {
return [];
}
const sourceCounts = []; // an array tohold the letter counts in source
const poolCounts = []; // an array tohold the letter counts in pool
// initialize counts for 26 letters to be 0
for (let i =0; i <26; i++) {
sourceCounts[i] =0;
poolCounts[i] =0;
}
// convert both strings to lower cases
pool = pool.toLocaleLowerCase();
const lowerSource = source.toLocaleLowerCase();
for (let i =0; i < pool.length; i++) {
// calculatepoolCounts for each letter in pool, mapping a - z to 0 - 25
poolCounts[pool[i].charCodeAt() -97]++;
}
const result = [];
for (let i =0; i < lowerSource.length; i++) {
// calculatesourceCounts for each letter for source, mapping a - z to 0 - 25
sourceCounts[lowerSource[i].charCodeAt() -97]++;
if (i >= pool.length-1) { // if source islong enough
// if sourceCountsis the same as poolCounts
if (JSON.stringify(sourceCounts) ===JSON.stringify(poolCounts)) {
// save the found anagram, using the original source to make stringcase-preserved
result.push(source.slice(i - pool.length+1, i +1));
}
// shift thestarting window by 1 index (drop the current first letter)
sourceCounts[lowerSource[i - pool.length+1].charCodeAt() -97]--;
}
}
// removeduplicates by a Set
return [...newSet(result)];
}

以下是验证测试:

console.log(showAnagrams( AaaAAaaAAaa ,  aa )); // ["Aa", "aa", "aA", "AA"]
console.log(showAnagrams( CbatobaTbacBoat ,  Boat ));  //["bato", "atob", "toba", "obaT","Boat"]
console.log(showAnagrams( AyaKkayakkAabkk ,  Kayak ));
// ["AyaKk", "yaKka", "aKkay", "Kkaya","kayak", "ayakk", "yakkA"]

回文

回文是从前往后读和从后往前读读法相同的单词或句子。有很多回文,比如A,Bob,还有 “A man, a plan, a canal — Panama”。检查回文的算法分为两种。使用循环或使用递归从两端检查是否相同。下列代码使用递归方法:

/**
* An algorithm to verify whethera given string is a palindrome
* @param {string} str The string to be verified
* @return {boolean} Returns whether it is a palindrome
*/
functionisPalindrome(str) {
functioncheckIsPalindrome(s) {
// empty stringor one letter is a defecto palindrome
if (s.length<2) {
returntrue;
}
if ( // if two ends notequal, ignoring the case
s[0].localeCompare(s[s.length-1], undefined, {
sensitivity:  base ,
}) !== 0
) {
returnfalse;
}
// since two ends equal, checking the inside
returncheckIsPalindrome(s.slice(1, -1));
}
// check whether it is a palindrome, removing noneletters and digits
returncheckIsPalindrome(str.replace(/[^A-Za-z0-9]/g,   ));
}

以下是验证测试:

console.log(isPalindrome( )); // true

console.log(isPalindrome( a )); // true

console.log(isPalindrome( Aa )); // true

console.log(isPalindrome( Bob )); // true

console.log(isPalindrome( Odd or even )); // false

console.log(isPalindrome( Never odd or even )); // true

console.log(isPalindrome( 02/02/2020 )); // true

console.log(isPalindrome( 2/20/2020 )); // false

console.log(isPalindrome( A man, a plan, a canal – Panama )); // true

回文面试题有很多不同的变形题,下面是一个在给定字符串中寻找最长回文的算法。

/**
* An algorithm to find thelongest palindrome in a given string
* @param {string} source The source to find the longest palindrome from
* @return {string} Returns the longest palindrome
*/
functionfindLongestPalindrome(source) {
// convert to lower cases and only keep lettersand digits
constlettersAndDigits = source.replace(/[^A-Za-z0-9]/g,   );
const str = lettersAndDigits.toLocaleLowerCase();
const len = str.length;
// empty string or one letter is a defecto palindrome
if (len <2) {
return str;
}
// the first letter is the current longest palindrome
let maxPalindrome = lettersAndDigits[0];
// assume that the index is the middle of a palindrome
for (let i =0; i < len; i++) {
// try the case that the palindrome has one middle
for (
let j =1; // start with onestep away (inclusive)
j < len &&// end with the len end (exclusive)
i - j >= 0&&// cannot pass the start index (inclusive)
i + j < len &&// cannot exceed end index (exclusive)
Math.min(2 * i +1, 2 * (len - i) -1) > maxPalindrome.length; // potential max length should be longer than thecurrent length
j++
) {
if (str[i - j] !== str[i + j]) { // if j stepsbefore the middle is different from j steps after the middle
break;
}
if (2 * j +1> maxPalindrome.length) { // if it is longerthan the current length
maxPalindrome = lettersAndDigits.slice(i - j, i + j +1); // j steps before, middle, and j steps after
}
}
// try the case that the palindrome has two middles
if (i < len -1&& str[i] === str[i +1]) { // if two middles are the same
if (maxPalindrome.length<2) { // the string withtwo middles could be the current longest palindrome
maxPalindrome = lettersAndDigits.slice(i, i +2);
}
for (
let j =1; // start with one step away (inclusive)
j < len -1&&// end with the len - 1 end (exclusive)
i - j >= 0&&// cannot pass the start index (inclusive)
i + j +1< len &&// cannot exceed end index (exclusive)
Math.min(2 * i +2, 2 * (len - i)) > maxPalindrome.length; // potential max length should be longer than thecurrent length
j++
) {
if (str[i - j] !== str[i + j +1]) { // if j stepsbefore the left middle is different from j steps after the right middle
break;
}
if (2 * j +2> maxPalindrome.length) { // if it is longer than the current length
maxPalindrome = lettersAndDigits.slice(i - j, i + j +2); // j steps before, middles, and j steps after
}
}
}
}
return maxPalindrome;
}

以下是验证测试:

console.log(findLongestPalindrome(  )); // ""
console.log(findLongestPalindrome( abc )); // "a"
console.log(findLongestPalindrome( Aabcd )); // "Aa"
console.log(findLongestPalindrome( I am Bob. )); // "Bob"
console.log(findLongestPalindrome( Odd or even )); // "Oddo"
console.log(findLongestPalindrome( Never odd or even )); // "Neveroddoreven"
console.log(findLongestPalindrome( Today is 02/02/2020. )); // "02022020"
console.log(findLongestPalindrome( It is 2/20/2020. )); // "20202"
console.log(findLongestPalindrome( A man, a plan, a canal – Panama )); // "AmanaplanacanalPanama"

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编译组:张月星、何婧璇

相关链接:

https://medium.com/better-programming/the-technical-interview-guide-to-string-manipulation-92f4c4649cd

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字符串操作大全:面试准备和日常编码所需一文打尽!

原文地址:https://blog.51cto.com/15057819/2564731

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