字符串有哪些操作方法
小编给大家分享一下字符串有哪些操作方法，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！
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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。
(1) 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}}(2) 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：
const map =newMap(); console.log(map.set(1, first )); // {1 =>"first"} console.log(map.set( a , second )); // {1 =>"first", "a" => "second"} console.log(map.set({ obj: 123 }, [1, 2, 3])); // {1 => "first", "a" =>"second", {obj: "123"} => [1, 2, 3]} console.log(map.set([2, 2, 2], newSet( abc ))); // {1 => "first", "a" => "second",{obj: "123"} => [1, 2, 3], [2, 2, 2] => {"a","b", "c"}} console.log(map.size); // 4 console.log(map.has(1)); // true console.log(map.get(1)); // "first" console.log(map.get( a )); // "second" console.log(map.get({ obj: 123 })); // undefined console.log(map.get([2, 2, 2])); // undefined console.log(map.delete(1)); // true console.log(map.has(1)); // false const arr = [3, 3]; map.set(arr, newSet( xyz )); console.log(map.get(arr)); // {"x", "y", "z"} console.log(map.keys()); // {"a", {obj: "123"}, [2, 2,2], [3, 3]} console.log(map.values()); // {"second", [1, 2, 3], {"a","b", "c"}, {"x", "y", "z"}} console.log(map.entries()); // {"a" => "second", {obj: "123"}=> [1, 2, 3], [2, 2, 2] => {"a", "b", "c"},[3, 3] => {"x", "y", "z"}} const map2 =newMap([[ a , 1], [ b , 2], [ c , 3]]); map2.forEach((value, key, map) => console.log(`value = ${value}, key = ${key}, map = ${map.size}`)); // value = 1, key = a, map = 3 // value = 2, key = b, map = 3 // value = 3, key = c, map = 3 map2.clear(); console.log(map2.entries()); // {}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。
(1) 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
(2) 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
(3) 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：
onsole.log(areIsomorphic( atlatl , tartar )); // true console.log(areIsomorphic( atlatlp , tartarq )); // true console.log(areIsomorphic( atlatlpb , tartarqc )); // true console.log(areIsomorphic( atlatlpa , tartarqb )); // false
(4) 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 poolpool = 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"]
(5) 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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