题目：原题链接（中等）
标签：字符串
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N ) O(N) O(N) | 416ms (13.98%) |
| Ans 2 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N ) O(N) O(N) | 44ms (65.68%) |
| Ans 3 (Python) |
解法一（暴力解法）：
class Solution:
def patternMatching(self, pattern: str, value: str) -> bool:
size1, size2 = len(pattern), len(value)
# 处理value为空的特殊情况
if size2 == 0:
return size1 == 1
for i1 in range(size2 + 1):
p1, s1 = pattern[0], value[:i1] # 模式值、对应的字符串值
j1 = 1 # 当前模式中的位置
i2 = i1 # 当前字符串中的位置
# 连续匹配当前值
right = True
while j1 < size1 and i2 < size2 and pattern[j1] == p1:
if value[i2:i2 + i1] == s1:
j1 += 1
i2 += i1
else:
right = False
break
# 处理已经匹配完成的情况
if j1 >= size1 or i2 >= size2:
if p1 not in pattern[j1:] and i2 == size2 and right:
return True
else:
continue
# 未能匹配则跳过
if not right:
continue
for i3 in range(i2 + 1, size2 + 1):
p2, s2 = pattern[j1], value[i2:i3]
j2 = j1 + 1
i4 = i3
# 检查当前情况是否正确
right = True
while j2 < size1 and i4 <= size2:
if pattern[j2] == p1:
if i1 > 0:
if value[i4:i4 + i1] == s1:
j2 += 1
i4 += i1
else:
right = False
break
else:
j2 += 1
else:
if i3 - i2 > 0:
if value[i4:i4 + (i3 - i2)] == s2:
j2 += 1
i4 += (i3 - i2)
else:
right = False
break
else:
j2 += 1
# print(right, "第1个:", s1, "第2个:", s2, "(", j2, "/", size1, ")", "(", i4, "/", size2, ")")
# 如果匹配成功则返回
if right and j2 == size1 and i4 == size2:
return True
return False
解法二：
class Solution:
def patternMatching(self, pattern: str, value: str) -> bool:
size1, size2 = len(pattern), len(value)
# 处理value为空的特殊情况
if size2 == 0:
return size1 == 1
count = collections.Counter(pattern)
count1, count2 = count["a"], count["b"]
# 如果有一个模式的出现频数为1，则令该模式为整串即可
if count1 == 1 or count2 == 1:
return True
# 处理只出现一种模式的情况
if count1 == 0 or count2 == 0:
if size2 % (count1 + count2) != 0:
return False
p = size2 // (count1 + count2)
for l in range(0, size2 - p, p):
if value[l:l + p] != value[l + p:l + 2 * p]:
return False
return True
# 处理两种模式均有出现的情况
for l1 in range(size2 // count1 + 1):
if (size2 - count1 * l1) % count2 != 0:
continue
l2 = (size2 - count1 * l1) // count2
a, b = "", ""
j1, j2 = 0, 0
while j1 < size1 and j2 < size2:
if pattern[j1] == "a":
if j2 + l1 > size2:
break
ch = value[j2:j2 + l1]
j2 += l1
if a == "":
a = ch
else:
if a != ch:
break
else:
if j2 + l2 > size2:
break
ch = value[j2:j2 + l2]
j2 += l2
if b == "":
b = ch
else:
if b != ch:
break
j1 += 1
else:
return True
return False
