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: falseExplanation: "a" does not match the entire string "aa".Example 2:Input:s = "aa"p = "a*"Output: trueExplanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".Example 3:Input:s = "ab"p = ".*"Output: trueExplanation: ".*" means "zero or more (*) of any character (.)".Example 4:Input:s = "aab"p = "c*a*b"Output: trueExplanation: 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:
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<!!!