10. Regular Expression Matching[H]正则表达式匹配

题目

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 . or * .

Example1：
Input：s = "aa" p="a"
Output：false
Explanation："a" does not match the entire string "a"
Example2：
Input：s = "aa" p="a*"
Output：true
Explanation："." means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it become "a".
Example3：
Input：s = "ab" p=".*"
Output：true
Explanation："." means " zero or more (*) of any character (.) " .

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Python知识

• 相同的值初始化(一维数组)

#方法一：list1 = [a a a ]
list1 = [ a for i in range(3)]
#方法二:
list1 = [a] * 3

• 二维数组初始化
  • 初始化一个**4*3**每项固定为0的数组

  list2 = [ [0 for i in range(3)] for j in range(4)]

  ---

  思路

  动态规划

  将待求解问题分解为若干个非互相独立的子问题，先求子问题，再求原问题。（通常需要将不同阶段的不同状态保存在二维数组内）

Step1.刻画一个最优解的结构特征：
\(dp[i][j]\)表示\(s[0,\cdots,i-1]\)与\(p[0,\cdots,j-1]\)是否匹配
Step2：递归定义最优解的值
1.\(p[j-1] == s [i-1]\)，则状态保存，\(dp[i][j] = dp[i-1][j-1]\)
2.\(p[j-1] ==\) .，.与任意单个字符匹配，于是状态保存，\(dp[i][j] = dp[i-1][j-1]\)
3.$p[j-1] == $*，*只能以X*的形式才能匹配，但是由于*究竟作为几个字符匹配不确定，此时有两种情况：

• \(p[j-2] != s[i-1]\)，此时\(s[0,\cdots,i-1]\)与\(p[0,\cdots,j-3]\)匹配，即\(dp[i][j] = dp[i][j-2]\)
• \(p[j-2] == s[i-1]\) 或者 $p[j-2] == \(`.`，此时应分为三种情况: `*`作为零个字符，\)dp[i][j] = dp[i][j-2]$
  *作为一个字符，\(dp[i][j] = dp[i][j-1]\)
  *作为多个字符，\(dp[i][j] = dp[i-1][j]\)

Step3.计算最优解的值
根据状态转移表，以及递推公式，计算dp[i][j]

---

C++

bool isMatch(string s, string p){
  int m = s.length(), n = p.length();
  bool dp[m+1][n+1];
  dp[0][0] = true; //空字符与空字符相匹配
  //初始化
  for(int i = 1; i <=m; i++)
    dp[i][0] = false;
  for(int i = 1; i <=n; i++)
    dp[0][i] = i > 1 && p[i-1] == '*' && dp[0][i-2];
  for(int i = 1; i<=m; i++){
    for(int j = 1; j<=n; j++){
      if(p[j-1] == '*'){
        dp[i][j] = dp[i][j-2] || (s[i-1] == p[j-2] || p[j-2] == '.') && dp[i-1][j];
      }
      else{
        dp[i][j] = (p[j-1] == '.' || s[i-1] == p[j-1]) && dp[i-1][j-1]
      }
    }
  }
}

---

Python

def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        len_s = len(s)
        len_p = len(p)
        dp = [[False for i in range(len_p+1)]for j in range(len_s+1)]
        dp[0][0] = True
        for i in range(1, len_p + 1):
            dp [0][i] = i>1 and dp[0][i - 2] and p[i-1] == '*'
        for i in range (1, len_s + 1 ):
            for j in range(1, len_p + 1):
                if p[j - 1] == '*':
                    #状态保留
                    dp[i][j] = dp[i][j -2] or (s[i-1] == p[j-2] or p[j-2] == '.') and dp[i-1][j]
                else:
                    dp[i][j] = (p[j-1] == '.' or s[i-1] == p[j-1]) and dp[i-1][j-1]
        return dp[len_s][len_p]