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Javascript 挑战:回文单词_Javascript - Fatal编程技术网

Javascript 挑战:回文单词

Javascript 挑战:回文单词,javascript,Javascript,回文是向后或向前读相同内容的单词。 编写一个函数,检查给定单词是否为回文。应忽略字符大小写。 函数isPalindromeword 例如,isAlindromedeleveled应返回true,因为应忽略字符大小写,从而导致deleveled,这是一个回文,因为它前后读取相同的内容 function isPalindrome(word) { // Please write your code here. } var word = readline() print(isPalindrome(w

回文是向后或向前读相同内容的单词。 编写一个函数,检查给定单词是否为回文。应忽略字符大小写。 函数isPalindromeword 例如,isAlindromedeleveled应返回true,因为应忽略字符大小写,从而导致deleveled,这是一个回文,因为它前后读取相同的内容

function isPalindrome(word)
{
  // Please write your code here.
}
var word = readline()
print(isPalindrome(word))
资料来源:

一种方法,链接上的其他方法:

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");
手动:

函数isPalindromeword{ word=word.toLowerCase; 对于变量i=0,j=word.length-1;iconsole.logisPalindromefoobar;这不是问问题,但它可能非常适合另一个Stack Exchange站点,例如,阅读他们的社区问题指南,然后首先搜索现有的问题。