# 10. Regular Expression Matching <span class='tag' data-diff='hard'></span>
{%hackmd RN5D4nggQRO8wzNqxuvlNw %}

## 題目
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 preceding 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

## 思路
這題如果用javascript內建的regular expression來做，則三行就解決了XDD

let re = new RegExp(p),
result = re.exec(s) || [];

return result[0] == s;

但題目應該是希望我們自己寫出parser。一開始想說把`s`跟`p`都先拆成array，再用迴圈不停地pop出最後一個元素，比較他們是否相同，若不同或是有一方空了則離開迴圈，並檢查他們是否都是空。

s = (["<s>"]).concat(s.split(""));
p = (["<s>"]).concat(p.split(""));

let a, b, flag = false;
while((a = flag ? a : p.pop()) != "<s>" && (b = s.pop()) != "<s>"){
    if(a == "*"){
        a = p.pop();
        flag = true;
    }
    if(a != b && a != "."){
        if(flag){
            s.push(b);
            flag = false;
            continue;
        }
        else {
            p.push(a);
            break;
        }
    }      
} 

其中`flag`的作用是檢查現在是否遇到`'*'`，若是則會把p再pop出一個元素作為比較的元素，並用該元素連續檢查s中pop出來的字元，直至不同，但這樣會有一個問題，如果遇到如下的測資:

s = "aaa"
p = "aa*"

則最末的`a*`就會將整個s比對完全，從而造成離開迴圈後`p`還有剩下的元素。