description:

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

my answer:

感恩

可以看出代码逻辑就是讨论pattern从0->2，
等于2时又分为第二个是不是无限匹配
最后还要处理一个‘’和p匹配的问题

大佬的answer：

class Solution {
public:
    bool isMatch(string s, string p) {
        if (p.empty()) return s.empty();
        if (p.size() == 1) {
            return (s.size() == 1 && (s[0] == p[0] || p[0] == '.'));
        }
        if (p[1] != '*') {
            if (s.empty()) return false;
            return (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p.substr(1));
            //这句话在写（抄）代码的时候报错了，根据控制变量法，最后判定错误出在&&，
            //因为我在模仿大佬的codestyle的时候&&右边打了两个空格（也可能是当时按错了输入法变       
            //成了全角emmmmm）
        }
        while (!s.empty() && (s[0] == p[0] || p[0] == '.')) {
            if (isMatch(s, p.substr(2))) return true;
            s = s.substr(1);
        }
        return isMatch(s, p.substr(2));
    }
};

relative point get√：

hint :

关键在于数清可能遇到的可能情况,就感觉easy考的都是基础的数据结构，越难就越考逻辑清晰，缜密，咋才能考虑到所有情况？--> 靠bug >o<!!!