[Leetcode]58. Length of Last Word

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58. Length of Last Word

題目連結
難易度：Easy

Given a string s consists of upper/lower-case alphabets and empty space characters ‘ ‘, return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example,

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Given s = "Hello World",
return 5.

找出給定一字串包含大小寫及空白字元，回傳最長的單字

思路

這題來試試看不熟悉的正規表達式，先定義出regex = /\S/g，\S其代表的意思是任何非空白鍵字元，若沒加g則只會匹配第一個找到的字元，
先用這個regex規則濾出沒有英文字的input， 使用match()來驗證，若返回null或input為空則return 0，接著使用trim()去除掉input的前後無效空白，
最後用split(' ')以空格拆解字串成陣列，回傳最後一個單字的長度即可。

解題

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/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLastWord = function(s) {
    var regex = /\S/g;
    if(s.match(regex) === null|| s=== '')  return 0;
    var result = s.trim().split(' ');
    return result[result.length-1].length;
};