代码思路:利用动态规划思想,设定dp[i][j]为s的前i个字符串与p的前j个字符串是否匹配,初始化第一行后进入动态规划循环,一般有下述几种情况:
- s[i-1]==p[j-1]
两字符相等,只需继承前一状态dp[i-1][j-1]即可 - s[i-1]!=p[j-1] and p[j-1]==’.’
两字符不等,且p当前字符为点,只需继承前一状态dp[i-1][j-1]即可 - s[i-1]!=p[j-1] and p[j-1]==’*’
两字符不等,且p当前字符为星,如果p上一字符是点或是值与s当前字符相等,则进行状态继承dp[i][j]=dp[i][j-1] or dp[i-1][j],同时,继承p上一字符状态dp[i][j]=dp[i][j] or dp[i][j-2]。
完成全部规划后输出结果。
class Solution:
def isMatch(self, s: str, p: str) -> bool:
m,n=len(s),len(p)
dp=[[False for i in range(n+1)] for i in range(m+1)]
dp[0][0]=True
for j in range (1,n+1):
if p[j-1]=='*':
dp[0][j]=dp[0][j-2]
else:dp[0][j]=False
for i in range (1,m+1):
for j in range(1,n+1):
if s[i-1]==p[j-1]:
dp[i][j]=dp[i-1][j-1]
elif p[j-1]=='.':
dp[i][j]=dp[i-1][j-1]
elif p[j-1]=='*':
if s[i-1]==p[j-2] or p[j-2]=='.':
dp[i][j]=dp[i][j-1] or dp[i-1][j]
dp[i][j]=dp[i][j] or dp[i][j-2]
return dp[m][n]