题目：原题链接（困难）
标签：字符串、哈希表、双指针、滑动窗口
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O(N×M×L2) : 其中M为words中词数，L为words中词长度 | O(M×L) | 704ms (61.09%) |
Ans 2 (Python) | O(N×L) : L为words中词长度 | O(M×L) | 112ms (81.47%) |
Ans 3 (Python) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一（暴力解法）：
def findSubstring(self, s: str, words: List[str]) -> List[int]:
# 处理特殊情况
if not s or not words:
return []
L = len(words[0])
total = len(words[0]) * len(words)
words = collections.Counter(words)
N = len(s)
ans = []
for i in range(N - total + 1):
tmp = [s[i + j:i + j + L] for j in range(0, total, L)]
for key, value in words.items():
if tmp.count(key) != value:
break
else:
ans.append(i)
return ans解法二（移动窗口）：
def findSubstring(self, s: str, words: List[str]) -> List[int]:
# 处理特殊情况
if not s or not words:
return []
L = len(words[0])
total = L * len(words)
words = collections.Counter(words)
N = len(s)
ans = []
for i in range(L):
left = i
right = i
now = collections.Counter()
while right < N:
if right - left < total:
now[s[right:right + L]] += 1
right += L
if right - left == total and now == words:
ans.append(left)
elif right - left == total:
now[s[left:left + L]] -= 1
if now[s[left:left + L]] == 0:
del now[s[left:left + L]]
now[s[right:right + L]] += 1
left += L
right += L
if now == words:
ans.append(left)
return ans
