[leetcode]Wildcard Matching @ Python

原题地址：https://oj.leetcode.com/problems/wildcard-matching/

题意：

Implement wildcard pattern matching with support for

'?'

and

'*'

.

'?' 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

解题思路：又是一个极其巧妙的解法。

Analysis:

For each element in s
If *s==*p or *p == ? which means this is a match, then goes to next element s++ p++.
If p=='*', this is also a match, but one or many chars may be available, so let us save this *'s position and the matched s position.
If not match, then we check if there is a * previously showed up,
if there is no *, return false;
if there is an *, we set current p to the next element of *, and set current s to the next saved s position.

e.g.

abed
?b*d**

a=?, go on, b=b, go on,
e=*, save * position star=3, save s position ss = 3, p++
e!=d, check if there was a *, yes, ss++, s=ss; p=star+1
d=d, go on, meet the end.
check the rest element in p, if all are *, true, else false;

Note that in char array, the last is NOT NULL, to check the end, use "*p" or "*p=='\0'".

代码：

class Solution:
# @param s, an input string
# @param p, a pattern string
# @return a boolean
# @good solution! use 'aaaabaaaab' vs 'a*b*b' as an example
def isMatch(self, s, p):
pPointer=sPointer=ss=0; star=-1
while sPointer<len(s):
if pPointer<len(p) and (s[sPointer]==p[pPointer] or p[pPointer]=='?'):
sPointer+=1; pPointer+=1
continue
if pPointer<len(p) and p[pPointer]=='*':
star=pPointer; pPointer+=1; ss=sPointer;
continue
if star!=-1:
pPointer=star+1; ss+=1; sPointer=ss
continue
return False
while pPointer<len(p) and p[pPointer]=='*':
pPointer+=1
if pPointer==len(p): return True
return False