leetcode刷题记录811-820 python版
前言
继续leetcode刷题生涯
这里记录的都是笔者觉得有点意思的做法
参考了好几位大佬的题解，感谢各位大佬
811. 子域名访问计数
class Solution:
def subdomainVisits(self, cpdomains: List[str]) -> List[str]:
if not cpdomains: return []
res = {}
for case in cpdomains:
time, web = case.split()
for i in range(len(web.split('.'))):
dm = web.split('.', i)[-1]
res[dm] = res.get(dm, 0) + int(time)
return [str(v) + ' ' + k for k, v in res.items()]
812. 最大三角形面积
from itertools import combinations
class Solution:
def largestTriangleArea(self, points: List[List[int]]) -> float:
return max(abs((x0-x2)*(y1-y2)-(x1-x2)*(y0-y2)) for (x0,y0),(x1,y1),(x2,y2) in combinations(points,3))/2
813. 最大平均值和的分组
class Solution:
def largestSumOfAverages(self, A: List[int], K: int) -> float:
N = len(A)
dp = [[0]*N for _ in range(K)]
for j in range(N-K+1):
dp[0][j] = sum(A[:j+1])/(j+1)
for i in range(1, K):
for j in range(i, N-K+i+1):
for p in range(i-1, j):
dp[i][j] = max(dp[i-1][p] + sum(A[p+1:j+1])/(j-p), dp[i][j])
return dp[-1][-1]
814. 二叉树剪枝
class Solution:
def pruneTree(self, root: TreeNode) -> TreeNode:
if not root: return None
root.left = self.pruneTree(root.left)
root.right = self.pruneTree(root.right)
if root.val == 0 and not root.left and not root.right:
return None
return root
815. 公交路线
class Solution:
def numBusesToDestination(self, routes: List[List[int]], S: int, T: int) -> int:
if S == T: return 0
d = {} #直达
for i in range(0, len(routes)):
for j in routes[i]:
if j in d:
d[j].add(i)
else:
d[j] = set({i})
h = set() #换乘
for key, value in d.items():
if len(value) > 1:
h.add(key)
visited = set({S})
queue = [S]
res = 1
n = 1
while len(queue) > 0:
front = queue.pop(0)
for bus in d[front]:
if T in routes[bus]:
return res
for site in routes[bus]:
if site in h and site not in visited:
queue.append(site)
visited.add(site)
n -= 1
if n == 0:
n = len(queue)
res += 1
return -1
816. 模糊坐标
import itertools
class Solution:
def ambiguousCoordinates(self, S: str) -> List[str]:
def zero(s):
if len(s) == 1:
return [s]
elif s[0] == '0' and s[-1] == '0':
return None
elif s[0] == '0':
return [s[0] + '.' + s[1:]]
elif s[-1] == '0':
return [s]
else:
return [s[:i]+'.'+s[i:] for i in range(1,len(s))] + [s]
S = S[1:-1]
n = len(S)
res = []
for i in range(1,n):
s1 = zero(S[:i])
s2 = zero(S[i:])
if s1 and s2:
for m, n in itertools.product(s1, s2):
res.append('(' + ', '.join([m,n]) + ')')
return res
# 笛卡尔积
class Solution:
def ambiguousCoordinates(self, S: str) -> List[str]:
def f(s):
if not s or len(s) > 1 and s[0] == s[-1] == '0':
return []
if len(s) > 1:
if s[0] == '0':
return [s[0] + '.' + s[1: ]]
if s[-1] == '0':
return [s]
return [s] + [s[: i] + '.' + s[i: ] for i in range(1, len(s))]
S = S[1: -1]
return ['(%s, %s)'%(x, y) for i in range(1, len(S)) for x, y in itertools.product(f(S[: i]), f(S[i: ]))]
817. 链表组件
class Solution:
def numComponents(self, head: ListNode, G: List[int]) -> int:
res = 0
tmp = flag = False
G_set = set(G)
while head:
flag = head.val in G_set
if (tmp ^ flag) & flag:
res += 1
tmp = flag
head = head.next
return res
818. 赛车
class Solution:
def racecar(self, target: int) -> int:
dp = [float('inf')] * (target + 1)
for i in range(1, target + 1):
k = 1 # A的次数
pos = 1
while pos < i: # pos < target
q = 0 # R过一次后A的次数
while ((1 << q) - 1) < pos:
dp[i] = min(dp[i], k + 1 + q + 1 + dp[i - (pos - ((1 << q) - 1))])
q += 1
k += 1
pos = (1 << k) - 1
if i == pos: # pos == target
dp[i] = k
else: # pos > target
dp[i] = min(dp[i], k + 1 + dp[pos - i])
return dp[target]
819. 最常见的单词
class Solution:
def mostCommonWord(self, paragraph: str, banned: List[str]) -> str:
import re
import collections
l = re.findall('[a-zA-Z]+',paragraph)
l = [i.lower() for i in l] #忽略大小写
count = collections.Counter(l) #建计数表
for i in count:
if i in banned:
count[i]=0 #ban词置0
res = count.most_common(1)
# res = list(count.items())
# res.sort(key=lambda x:x[1])
return res[0][0]
820. 单词的压缩编码
class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
word = set(words)
sum = 0
for i in words:
for j in range(1, len(i)):
word.discard(i[j:])
for k in word:
sum += len(k)+1
return sum