检查JavaScript中回文的两种方法
本文基于Free Code Camp基本算法脚本“ Palindromes检查”。
palindrome是一个单词，词组，数字或其他顺序向后或向前读取的字符。“palindrome”一词最早是由英国剧作家本·琼森（Ben Jonson）在17世纪创造的，其起源是希腊语的palin（“again”）和dromos（“ way，direction”）。— src.维基百科
在本文中，我将解释两种方法，一种是使用内置函数，另一种是使用for循环。
算法挑战
如果给定的字符串是回文，则返回true。否则，返回false。
回文是指向前或向后以相同方式拼写的单词或句子，而忽略了标点，大小写和空格。
注意：您需要删除所有非字母数字字符（标点，空格和符号），并将所有内容都转换为小写以检查palindrome。
我们将传递各种格式的字符串，例如“ racecar”，“ RaceCar”和“ race CAR”。
function palindrome(str) {
return true;
}
palindrome("eye");
提供的测试用例
- palindrome(“race car”)返回true
- palindrome(“not a palindrome”)返回false
- palindrome(“A man, a plan, a canal. Panama”)返回true
- palindrome(“neverodd or even”)返回true
- palindrome(“nope”)返回false
- palindrome(“almostomla”)返回false
- palindrome(“My age is 0, 0 si ega ym.”)返回true
- palindrome(“1 eye for of 1 eye.”)返回false
- palindrome(“0_0(: /-\ :） 0–0”)返回true
我们需要哪个正确表达式通过最后一个测试用例？
正确表达式是用于匹配字符串中字符组合的模式。
当搜索匹配项不仅仅需要直接匹配项时，该模式将包含特殊字符。
To pass the last test case, we can use two Regular Expressions:
/[^A-Za-z0–9]/g or
/[\W_]/g
\ W删除所有非字母数字字符：
- \ W匹配任何非单词字符
- \ W等效于[^ A-Za-z0–9_]
- \ W匹配括号中未包含的任何内容
这意味着什么？
[^A-Z] matches anything that is not enclosed between A and Z
[^a-z] matches anything that is not enclosed between a and z
[^0-9] matches anything that is not enclosed between 0 and 9
[^_] matches anything that does not enclose _
但是在我们的测试案例中，我们需要palindrome（“ 0_0（：/-\ :）0–0 ”）返回true，这意味着必须匹配“ _（：/-\：）– ”。
我们将需要添加“ _ ”以通过此特定测试用例。
We now have “\W_”
我们还需要添加g标志进行全局搜索。
We finally have “/[\W_]/g”
- / [\ W _] / g用于纯粹的演示目的，以显示RegExp的工作方式。/ [^ A-Za-z0–9] / g是最容易选择的RegExp 。
1.检查具有内置功能的回文
对于此解决方案，我们将使用几种方法：
- toLowerCase（）的方法返回调用字符串值转换为小写。
- replace()的方法返回的部分或由替换替换模式的所有匹配一个新的字符串。我们将使用我们刚才创建的RegExp之一。
- split()的方法通过分离串分成子串分割字符串对象到字符串数组。
- reverse()的方法在当前位置倒转的阵列。第一个数组元素变为最后一个，最后一个数组变为第一个。
- join()的方法连接到一个字符串数组的所有元素。
function palindrome(str) {
// Step 1. Lowercase the string and use the RegExp to remove unwanted characters from it
var re = /[\W_]/g; // or var re = /[^A-Za-z0-9]/g;
var lowRegStr = str.toLowerCase().replace(re, '');
// str.toLowerCase() = "A man, a plan, a canal. Panama".toLowerCase() = "a man, a plan, a canal. panama"
// str.replace(/[\W_]/g, '') = "a man, a plan, a canal. panama".replace(/[\W_]/g, '') = "amanaplanacanalpanama"
// var lowRegStr = "amanaplanacanalpanama";
// Step 2. Use the same chaining methods with built-in functions from the previous article 'Three Ways to Reverse a String in JavaScript'
var reverseStr = lowRegStr.split('').reverse().join('');
// lowRegStr.split('') = "amanaplanacanalpanama".split('') = ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"]
// ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"].reverse() = ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"]
// ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"].join('') = "amanaplanacanalpanama"
// So, "amanaplanacanalpanama".split('').reverse().join('') = "amanaplanacanalpanama";
// And, var reverseStr = "amanaplanacanalpanama";
// Step 3. Check if reverseStr is strictly equals to lowRegStr and return a Boolean
return reverseStr === lowRegStr; // "amanaplanacanalpanama" === "amanaplanacanalpanama"? => true
}
palindrome("A man, a plan, a canal. Panama");
没有注释：
function palindrome(str) {
var re = /[\W_]/g;
var lowRegStr = str.toLowerCase().replace(re, '');
var reverseStr = lowRegStr.split('').reverse().join('');
return reverseStr === lowRegStr;
}
palindrome("A man, a plan, a canal. Panama");
2.用FOR循环检查回文
半索引（len / 2）在处理大字符串时有好处。我们检查每个部分的结尾，然后将FOR循环内的迭代次数除以2。
function palindrome(str) {
// Step 1. The first part is the same as earlier
var re = /[^A-Za-z0-9]/g; // or var re = /[\W_]/g;
str = str.toLowerCase().replace(re, '');
// Step 2. Create the FOR loop
var len = str.length; // var len = "A man, a plan, a canal. Panama".length = 30
for (var i = 0; i < len/2; i++) {
if (str[i] !== str[len - 1 - i]) {
// As long as the characters from each part match, the FOR loop will go on
return false; // When the characters don't match anymore, false is returned and we exit the FOR loop
}
/* Here len/2 = 15
For each iteration: i = ? i < len/2 i++ if(str[i] !== str[len - 1 - i])?
1st iteration: 0 yes 1 if(str[0] !== str[15 - 1 - 0])? => if("a" !== "a")? // false
2nd iteration: 1 yes 2 if(str[1] !== str[15 - 1 - 1])? => if("m" !== "m")? // false
3rd iteration: 2 yes 3 if(str[2] !== str[15 - 1 - 2])? => if("a" !== "a")? // false
4th iteration: 3 yes 4 if(str[3] !== str[15 - 1 - 3])? => if("n" !== "n")? // false
5th iteration: 4 yes 5 if(str[4] !== str[15 - 1 - 4])? => if("a" !== "a")? // false
6th iteration: 5 yes 6 if(str[5] !== str[15 - 1 - 5])? => if("p" !== "p")? // false
7th iteration: 6 yes 7 if(str[6] !== str[15 - 1 - 6])? => if("l" !== "l")? // false
8th iteration: 7 yes 8 if(str[7] !== str[15 - 1 - 7])? => if("a" !== "a")? // false
9th iteration: 8 yes 9 if(str[8] !== str[15 - 1 - 8])? => if("n" !== "n")? // false
10th iteration: 9 yes 10 if(str[9] !== str[15 - 1 - 9])? => if("a" !== "a")? // false
11th iteration: 10 yes 11 if(str[10] !== str[15 - 1 - 10])? => if("c" !== "c")? // false
12th iteration: 11 yes 12 if(str[11] !== str[15 - 1 - 11])? => if("a" !== "a")? // false
13th iteration: 12 yes 13 if(str[12] !== str[15 - 1 - 12])? => if("n" !== "n")? // false
14th iteration: 13 yes 14 if(str[13] !== str[15 - 1 - 13])? => if("a" !== "a")? // false
15th iteration: 14 yes 15 if(str[14] !== str[15 - 1 - 14])? => if("l" !== "l")? // false
16th iteration: 15 no
End of the FOR Loop*/
}
return true; // Both parts are strictly equal, it returns true => The string is a palindrome
}
palindrome("A man, a plan, a canal. Panama");
没有注释：
function palindrome(str) {
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
var len = str.length;
for (var i = 0; i < len/2; i++) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
palindrome("A man, a plan, a canal. Panama");
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