[Leetcode][Python]44：Wildcard Matching
# -*- coding: utf8 -*-'''
__author__ = 'dabay.wang@gmail.com'
44：Wildcard Matching
https://oj.leetcode.com/problems/wildcard-matching/
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false
===Comments by Dabay===
使用动态规划的做法.
先想象在s和p的前面都加一个空格，以s为外循环，p为内循环。
先初始化第一组数据，第一个“空”匹配“空”，所以第一个为True。然后看如果后面如果是*就继续设置为True。
进入循环之后，判断根据p中的字符
- 如果是？*以外的字符，就判断这个字符和s中对应的字符是否一致，而且s中到上一个字符和p中到上一次字符匹配，设True
- 如果是？，如果s中到上一个字符和p中到上一个字符匹配，设True
- 如果是*，满足下面的情况，就设True
- （匹配0次）s中到这个字符和p到上两个字符匹配
- （匹配1次）s中到这个字符和p到上一个字符匹配
- （匹配n次）s中到上一个字符和p到这个字符匹配
'''
class Solution:
# @param s, an input string
# @param p, a pattern string
# @return a boolean
def isMatch(self, s, p):
def quick_test(s, p):
num_of_star = 0
for x in p:
if x == "*":
num_of_star = num_of_star + 1
return len(s) >= len(p)-num_of_star
if quick_test(s, p) is False:
return False
default_row = +
current_row = list(default_row)
current_row = True
for j in xrange(len(p)):
if p == "*":
current_row = True
else:
break
previous_row = current_row
for i in xrange(len(s)):
current_row = list(default_row)
for j in xrange(len(p)):
if p == "?":
if previous_row:
current_row = True
elif p == "*":
if current_row or previous_row or previous_row:
current_row = True
else:
if p == s and previous_row:
current_row = True
previous_row = current_row
return previous_row[-1]
def main():
sol = Solution()
s = "ac"
p = "ab*"
print sol.isMatch(s, p)
if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)
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