59 lines
1.4 KiB
Python
59 lines
1.4 KiB
Python
#coding: utf-8
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''' mbinary
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#######################################################################
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# File : wildcard_matching.py
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# Author: mbinary
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# Mail: zhuheqin1@gmail.com
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# Blog: https://mbinary.xyz
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# Github: https://github.com/mbinary
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# Created Time: 2018-12-13 22:46
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# Description:
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wild card '*' matches 0 or any chars, and '?' matches any single char.
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#######################################################################
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'''
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'''
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idea
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dynamic programming
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dp[m+1][n+1]: bool
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i:n, j:m
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dp[j][i] indicates if s[:i+1] matches p[:j+1]
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initial: dp[0][0] = True, dp[0][i],dp[j][0] = False
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only if p startswith '*', dp[1][0] = True.
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if p[j] = '*': dp[j][i] = dp[j-1][i] or dp[j][i-1]
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elif p[j] = '?': dp[j][i] = dp[j-1][i-1]
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else : dp[j][i] = dp[j-1][i-1] and s[i] == p[j]
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'''
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# leetcode: q44 https://leetcode.com/problems/wildcard-matching/description/
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def isMatch(self, s, p):
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"""
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:type s: str
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:type p: str pattern str including wildcard
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:rtype: bool
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"""
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n, m = len(s), len(p)
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last = [False]*(n+1)
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last[0] = True
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for j in range(m):
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if p[j] == '*':
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for i in range(n):
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last[i+1] = last[i+1] or last[i]
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elif p[j] == '?':
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last.pop()
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last.insert(0, False)
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else:
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li = [False]
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for i in range(n):
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li.append(last[i] and p[j] == s[i])
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last = li
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return last[-1]
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