题目链接

说明

• 以p= a\*b\*, s = abbb 为例
• 下图为dp的矩阵，第一列和第一行用于帮助后续的状态转移，无实际含义

1. rows = len(s)+1, cols = len(p)+1
2. dp[rows][cols], dp[0][0] = True
3. dp的第0行要初始化，为了后续的状态转移: 找到模式串的第一个*，此处的dp状态与* 的前面两个字符的dp状态相同

   for i in range(1, cols):
     if p[i-1] == '*':
         dp[0][i] = dp[0][i-2]
   

思路

1. s[i-1] == p[j-1] or p[j-1] == '.' 则: dp[i][j] = dp[i-1][j-1]
2. p[j-1] == '*' 则:
   • if s[i-1] != p[j-2] : dp[i][j] = dp[i][j-2]
   • if s[i-1] == p[j-2] or p[j-2] == '.' : dp[i][j] = dp[i-1][j] or dp[i][j-1]

代码实现

	class Solution:
      def isMatch(self, s: str, p: str) -> bool:
          rows = len(s)+1
          cols = len(p)+1
          dp = [[False for _ in range(cols)] for _ in range(rows)]
          dp[0][0] = True
          for i in range(1, cols):
              if p[i-1] == '*':
                  dp[0][i] = dp[0][i-2]
          for i in range(1, rows):
              for j in range(1, cols):
                  if s[i-1] == p[j-1] or p[j-1] == '.':
                      dp[i][j] = dp[i-1][j-1]
                  elif p[j-1] == '*':
                      if s[i-1] != p[j-2] and p[j-2] != '.':
                          dp[i][j] = dp[i][j-2]
                      else: 
                          if s[i-1] == p[j-2] or p[j-2] == '.':        
                              dp[i][j] = dp[i-1][j] or dp[i][j-1]
                          dp[i][j] |= dp[i][j-2]
                  else: dp[i][j] = False
          return dp[rows-1][cols-1]