Given an input string (`s`) and a pattern (`p`), implement regular expression matching with support for`'.'`and`'*'`. ~~~ '.' Matches any single character. '*' Matches zero or more of the preceding element. ~~~ The matching should cover the**entire**input string (not partial). **Note:** * `s`?could be empty and contains only lowercase letters`a-z`. * `p`could be empty and contains only lowercase letters`a-z`, and characters like?`.`?or?`*`. **Example 1:** ~~~ Input: s = "aa" p = "a" Output: false Explanation: "a" does not match the entire string "aa". ~~~ **Example 2:** ~~~ Input: s = "aa" p = "a*" Output: true Explanation:?'*' means zero or more of the precedeng?element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". ~~~ **Example 3:** ~~~ Input: s = "ab" p = ".*" Output: true Explanation:?".*" means "zero or more (*) of any character (.)". ~~~ **Example 4:** ~~~ Input: s = "aab" p = "c*a*b" Output: true Explanation:?c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab". ~~~ **Example 5:** ~~~ Input: s = "mississippi" p = "mis*is*p*." Output: false ~~~ ``` /** * @param {string} s * @param {string} p * @return {boolean} */ var isMatch = function(s, p) { if(p == '' || !p){ return s == ''; } // 判斷 s 是否為空 防止越界，如果 s 為空， // 表達式直接為 false , s.charAt(0) 就不會執行了 var first_match = (s && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.' )); // 當長度大于2的時候，才考慮 * // ---兩種情況 //p 直接跳過兩個字符，表示 * 前邊的字符出現 0 次 //p 不變，接著用第一個字符和前面比較， // 表示 * 用前一個字符替代 【s = aa , p = a*】 if(p.length >= 2 && p.charAt(1) == '*'){ return ( isMatch(s,p.substring(2)) || (first_match && isMatch(s.substring(1),p)) ) }else { return first_match && isMatch(s.substring(1),p.substring(1)); } }; ```