"""
題目：給你乙個字串s、乙個字串t，請在字串s裡面找出：包含t所有字母的最小子串。
示例：輸入：s="adobecodebanc", t="abc
輸出："banc"
"""# 方法一 自己寫的，但是leetcode上超時
defminwindow
(s, t)
: m =
0 all_in =
false
iflen
(t)==0or
len(s)==0
:return
""if
len(t)
>
len(s)
:return
"" minlength =
len(t)
str=
while
true
:for i in
range
(len
(s)-minlength+1)
: left = i
right = i + minlength
for k in s[left:right]
:str
for j in
range
(len
(t))
:for n in
range
(len
(str))
:if t[j]
==str
[n]:
m +=
1str
.pop(n)
break
if m ==
len(t)
: all_in =
true
str=
m =0if all_in:
return s[left:right]
minlength +=
1if minlength ==
len(s)+1
:return
""print
(minwindow(
"adobecodebanc"
,"r"))
print
(minwindow(
"acbbaca"
,"aba"))
# 方法二 leetcode上的官方解法
from collections import counter
defminwindow
(s, t):if
not t or
not s:
return
"" dict_t = counter(t)
required =
len(dict_t)
# t中不重複元素的個數
filtered_s =
# t中元素在s**現的情況(位置, 值)
for i, char in
enumerate
(s):
if char in dict_t:
(i, char)
) l, r =0,
0 formed =
0 window_counts =
ans =
float
("inf"),
none
,none
while r <
len(filtered_s)
: character = filtered_s[r][1
] window_counts[character]
= window_counts.get(character,0)
+1if window_counts[character]
== dict_t[character]
: formed +=
1# 直到t**現的元素在滑動視窗中全有， 否則一直擴大滑動視窗
while l <= r and formed == required:
# 判斷視窗是否可以縮小，left右移
character = filtered_s[l][1
] end = filtered_s[r][0
] start = filtered_s[l][0
]if end - start +
1< ans[0]
: ans =
(end - start +
1, start, end)
window_counts[character]-=1
if window_counts[character]
< dict_t[character]
: formed -=
1 l +=
1 r +=
1return
""if ans[0]
==float
("inf"
)else s[ans[1]
:ans[2]
+1]# print(minwindow("adobecodebanc", "abc"))
print
(minwindow(
"acbbccd"
,"ad"
))
76 最小覆蓋子串
給你乙個字串 s 乙個字串 t，請在字串 s 裡面找出 包含 t 所有字母的最小子串。示例 輸入 s adobecodebanc t abc 輸出 banc 說明 如果 s 中不存這樣的子串，則返回空字串 如果 s 中存在這樣的子串，我們保證它是唯一的答案。class solution int ma...
76最小覆蓋子串
給你乙個字串 s 乙個字串 t，請在字串 s 裡面找出 包含 t 所有字母的最小子串。示例 輸入 s adobecodebanc t abc 輸出 banc 說明 滑動視窗解法。hashmap方式 param s param t return public static string minwind...
76 最小覆蓋子串
給你乙個字串 s 乙個字串 t，請在字串 s 裡面找出 包含 t 所有字元的最小子串。示例 輸入 s adobecodebanc t abc 輸出 banc 說明 如果 s 中不存這樣的子串，則返回空字串 如果 s 中存在這樣的子串，我們保證它是唯一的答案。使用滑動視窗的思想，我們可以用乙個map來...