Description:
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
算法分析:“*”把p分成若干部分,然後判斷,這幾部分是不是可以能在s中匹配到。先定義兩個指針從s和p的頭出發如果p的第一個部分可以和s的前面部分匹配,那麼兩個指針往右移同樣長度到第二部分,直到找到最後。例如,如果s爲“abba”,p爲“a*a”。首先,我們判斷*前面那部分是不是和s的前面部分匹配。’a‘和‘a‘可以匹配;那麼判斷p中第二個‘a‘在s中能否匹配。發現可以,所以 返回true。這個的時間複雜度是(O(n))。
代碼如下:
class Solution(object):
def match(self, s, p, i, j, size):
k = 0
while k < size:
if s[k + i] != p[j + k] and p[j + k] != '?':
return False
k += 1
return True
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
if s == p or p == "*":
return True
size = len(p);
d = []
if size == 0 or len(s) == 0:
return False
for i in range(size):
if p[i] == '*':
d.append(i)
i = 0;
j = 0;
k = 0;
ss = len(s)
if len(d) == 0:
if ss != size:
return False
return self.match(s, p, 0, 0, size)
first = True
while i < ss:
if ss - i < size - j - len(d) + k:
return False
if self.match(s, p, i, j, d[k] - j):
first = False
if k == len(d) - 1:
if d[k] == size - 1:
return True
tmp = size - 1 - d[k]
return self.match(s, p, ss - tmp, d[k] + 1, tmp)
i += d[k] - j;
j = d[k] + 1;
k += 1
if i == ss:
while j < size:
if p[j] != '*':
return False
j += 1
return True
elif first:
return False
else:
i += 1
return False
運行結果: