LeetCode 每日一题 2025/2/24-2025/3/2
记录了初步解题思路 以及本地实现代码;并不一定为最优 也希望大家能一起探讨 一起进步
目录
- 2/24 1656. 设计有序流
- 2/25 2502. 设计内存分配器
- 2/26 1472. 设计浏览器历史记录
- 2/27 2296. 设计一个文本编辑器
- 2/28 2353. 设计食物评分系统
- 3/1 131. 分割回文串
- 3/2
2/24 1656. 设计有序流
数组保存数据 插入数据后
检查指针节点是否可以往后移动
class OrderedStream(object):
def __init__(self, n):
"""
:type n: int
"""
self.n = n
self.ptr = 0
self.l = [""]*(n+1)
def insert(self, idKey, value):
"""
:type idKey: int
:type value: str
:rtype: List[str]
"""
ans = []
self.l[idKey] = value
while self.ptr<self.n and self.l[self.ptr+1]!="":
ans.append(self.l[self.ptr+1])
self.ptr +=1
if self.ptr>self.n:
break
return ans
2/25 2502. 设计内存分配器
使用长度为n的数字 记录每个位置内存使用情况
class Allocator(object):
def __init__(self, n):
"""
:type n: int
"""
self.n=n
self.mem=[0]*n
def allocate(self, size, mID):
"""
:type size: int
:type mID: int
:rtype: int
"""
cnt = 0
for i in range(self.n):
if self.mem[i]>0:
cnt=0
else:
cnt+=1
if cnt==size:
for j in range(i-cnt+1,i+1):
self.mem[j]=mID
return i-cnt+1
return -1
def freeMemory(self, mID):
"""
:type mID: int
:rtype: int
"""
cnt = 0
for i in range(self.n):
if self.mem[i]==mID:
cnt+=1
self.mem[i]=0
return cnt
2/26 1472. 设计浏览器历史记录
模拟 使用一个数组urls保存网页记录
cur指向当前访问的网页索引
class BrowserHistory(object):
def __init__(self, homepage):
"""
:type homepage: str
"""
self.urls = [homepage]
self.cur = 0
def visit(self, url):
"""
:type url: str
:rtype: None
"""
self.urls = self.urls[:self.cur+1]
self.urls.append(url)
self.cur+=1
def back(self, steps):
"""
:type steps: int
:rtype: str
"""
self.cur = max(0,self.cur-steps)
return self.urls[self.cur]
def forward(self, steps):
"""
:type steps: int
:rtype: str
"""
self.cur = min(len(self.urls)-1,self.cur+steps)
return self.urls[self.cur]
2/27 2296. 设计一个文本编辑器
使用left,right两个栈分别存储光标左右的内容
add:将text压入left中
delete:从left中取出k个直至为空
cursorleft:将left中取出放入right中
cursorright:将right中取出放入left中
class TextEditor(object):
def __init__(self):
self.left=[]
self.right=[]
def addText(self, text):
"""
:type text: str
:rtype: None
"""
self.left.extend(text)
def deleteText(self, k):
"""
:type k: int
:rtype: int
"""
cnt=min(k,len(self.left))
del self.left[-cnt:]
return cnt
def cursorLeft(self, k):
"""
:type k: int
:rtype: str
"""
for _ in range(min(k,len(self.left))):
self.right.append(self.left.pop())
return ''.join(self.left[-10:])
def cursorRight(self, k):
"""
:type k: int
:rtype: str
"""
for _ in range(min(k,len(self.right))):
self.left.append(self.right.pop())
return ''.join(self.left[-10:])
2/28 2353. 设计食物评分系统
foodmap维护food对应的分数和烹饪方法
ratemap[cuisines] 使用最小堆维护同一烹饪方法中的分数 用负数变为最大堆
import heapq
class FoodRatings(object):
def __init__(self, foods, cuisines, ratings):
"""
:type foods: List[str]
:type cuisines: List[str]
:type ratings: List[int]
"""
self.food={}
self.rate={}
self.n=len(foods)
for i in range(self.n):
f = foods[i]
c = cuisines[i]
r = ratings[i]
self.food[f]=(c,r)
if c not in self.rate:
self.rate[c]=[]
heapq.heappush(self.rate[c], (-r,f))
def changeRating(self, food, newRating):
"""
:type food: str
:type newRating: int
:rtype: None
"""
c,r = self.food[food]
heapq.heappush(self.rate[c], (-newRating,food))
self.food[food]=(c,newRating)
def highestRated(self, cuisine):
"""
:type cuisine: str
:rtype: str
"""
while self.rate[cuisine]:
r,f = self.rate[cuisine][0]
if -r == self.food[f][1]:
return f
heapq.heappop(self.rate[cuisine])
return ""
3/1 131. 分割回文串
check检查字符串l是否回文
pdic存放pos开头所有回文的子串
def partition(s):
"""
:type s: str
:rtype: List[List[str]]
"""
def check(l):
t = True
if len(l)==0:
return t
while t and len(l)>1:
if l[0]==l[-1]:
l.pop(0)
l.pop(-1)
else:
t = False
if len(l)>1:
return False
else:
return True
ret =[]
if len(s)==0:
return ret
l = list(s)
dic = {}
for i in range(len(l)):
begin = l[i]
for j in range(len(l)-1,i-1,-1):
if begin == l[j]:
if check(l[i:j+1]):
dic[(i,j)] = l[i:j+1]
pdic = {}
for i in dic:
pos = i[0]
tmp = pdic.get(pos,[])
tmp.append(i)
pdic[pos] = tmp
def combine(begin):
ret = []
va = pdic[begin]
for v in va:
end = v[1]
if end == (len(s)-1):
ret.append([v])
else:
l = combine(end+1)
for t in l:
t.insert(0,v)
ret.extend(l)
return ret
r = combine(0)
for v in r:
tmp = []
for i in v:
tmp.append(''.join(dic[i]))
ret.append(tmp)
return ret
3/2