字符串有哪些操作方法 - web开发
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!
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。
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dermatoglyphics (15个字母)
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hydropneumatics (15个字母)
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misconjugatedly (15个字母)
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uncopyrightable (15个字母)
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uncopyrightables (16个字母)
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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:
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Waltz, bad nymph, for quick jigs vex. (28个字母)
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Jived fox nymph grabs quick waltz. (28个字母)
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Glib jocks quiz nymph to vex dwarf. (28个字母)
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Sphinx of black quartz, judge my vow. (29个字母)
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How vexingly quick daft zebras jump! (30个字母)
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The five boxing wizards jump quickly. (31个字母)
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Jackdaws love my big sphinx of quartz. (31个字母)
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Pack my box with five dozen liquor jugs. (32个字母)
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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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发布于:2023-01-18,除非注明,否则均为
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