给你一个字符串 s 、一个字符串 t 。返回 s 中涵盖 t 所有字符的最小子串。如果 s 中不存在涵盖 t 所有字符的子串,则返回空字符串 "" 。
注意:
对于 t 中重复字符,我们寻找的子字符串中该字符数量必须不少于 t 中该字符数量。
如果 s 中存在这样的子串,我们保证它是唯一的答案。
示例 1:
输入:s = "ADOBECODEBANC", t = "ABC"
输出:"BANC"
示例 2:
输入:s = "a", t = "a"
输出:"a"
示例 3:
输入: s = "a", t = "aa"
输出: ""
解释: t 中两个字符 'a' 均应包含在 s 的子串中,
因此没有符合条件的子字符串,返回空字符串。
提示:
1 <= s.length, t.length <= 105
s 和 t 由英文字母组成
进阶:你能设计一个在 o(n) 时间内解决此问题的算法吗?
# @lc app=leetcode.cn id=76 lang=python3
#
# [76] 最小覆盖子串
#
#滑动窗口方法
# @lc code=start
class Solution:
def minWindow(self, s: str, t: str) -> str:
dict1=dict()
for i in t:
if i in dict1:
dict1[i]+=1
else:
dict1[i]=1
def isempty(dicts):
for i in dicts:
if dicts[i]>0:
return False
return True
if len(t)==1:
if t in s:return t
else: return ""
j=-1
i=0
m=i
n=j
lens= 10**5
while i<=len(s)-len(t):
if isempty(dict1):
if j-i+1<lens:
lens=j-i+1
m=i
n=j
if s[i] in dict1:
dict1[s[i]]= dict1[s[i]]+1
i=i+1
elif j<len(s)-1:
j=j+1
if s[j] in dict1 :
dict1[s[j]]=dict1[s[j]]-1
elif j==len(s)-1:
i=i+1
if lens == 10**5:
return ""
else: return s[m:n+1]
# @lc app=leetcode.cn id=73 lang=python3
#
# [73] 矩阵置零
#
# @lc code=start
class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
m=len(matrix)
n=len(matrix[0])
res=list()
for i in range(m):
for j in range(n):
if matrix[i][j]==0:
res.append([i,j])
for l in res:
matrix[l[0]]=[0]*n
for x in matrix:
x[l[1]]=0
输入:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
输出:
[
"This is an",
"example of text",
"justification. "
]
示例 2:
输入:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
输出:
[
"What must be",
"acknowledgment ",
"shall be "
]
解释: 注意最后一行的格式应为 "shall be " 而不是 "shall be",
因为最后一行应为左对齐,而不是左右两端对齐。
第二行同样为左对齐,这是因为这行只包含一个单词。
示例 3:
输入:
words = ["Science","is","what","we","understand","well","enough","to","explain",
"to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
输出:
[
"Science is what we",
"understand well",
"enough to explain to",
"a computer. Art is",
"everything else we",
"do "
]
code:
# [68] 文本左右对齐
#
# @lc code=start
class Solution:
def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
lens=len(words)
rev=list()
nowlens=0
start=0
i=0
while i<lens:
wordlen=len(words[i])
nowlens=nowlens+wordlen
#print(i,nowlens)
if nowlens<maxWidth:
nowlens+=1
i=i+1
#print(i)
#print(nowlens)
elif nowlens>maxWidth:
nowlens=nowlens-wordlen-1
reslens=maxWidth-nowlens
m=(i-1-start)
if m==0:
mid=""
mid=mid+words[start]+(maxWidth-len(words[start]))*" "
rev.append(mid)
#print(reslens,reslens//(i-1-start),reslens/(i-1-start))
elif reslens//(i-1-start)==reslens/(i-1-start):
mid=""
x=reslens//(i-1-start)+1
for j in range(start,i-1):
mid=mid+words[j]+x*" "
mid=mid+words[i-1]
rev.append(mid)
else:
other=reslens%(i-1-start)
mid=""
x=reslens//(i-1-start)+1
print(x,other)
for j in range(start,i-1):
if other>0:
mid=mid+words[j]+x*" "+" "
other-=1
else:
mid=mid+words[j]+x*" "
mid=mid+words[i-1]
rev.append(mid)
nowlens=0
start=i
elif nowlens==maxWidth:
mid=""
for j in range(start,i):
mid=mid+words[j]+" "
mid=mid+words[i]
rev.append(mid)
start=i+1
nowlens=0
i=i+1
if start<lens:
mid=""
for j in range(start,lens):
mid=mid+words[j]+" "
mid=mid[:-1]
other=maxWidth-len(mid)
mid=mid+other*" "
rev.append(mid)
return rev
# @lc code=end