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

题目

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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 (.) " .

思路

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动态规划

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]

Tips

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数组初始化（python）

（1）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

（2）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)]

C++

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(),n = p.length();
        bool dp[m+1][n+1];
        dp[0][0] = true;
        //初始化第0行,除了[0][0]全为false，因为空串p只能匹配空串，其他都无能匹配
        for (int i = 1; i <= m; i++)
            dp[i][0] = false;
        //初始化第0列，只有X*能匹配空串
        for (int j = 1; j <= n; j++)
            dp[0][j] = j > 1 && '*' == p[j - 1] && dp[0][j - 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 //只有当前字符完全匹配，才能传递dp[i-1][j-1] 值
               {
                   dp[i][j] = (p[j - 1] == '.' || s[i - 1] == p[j - 1]) && dp[i - 1][j - 1];
               }
           }
        }
        return dp[m][n];
    }
};

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]