LeetCode 每日一题 2023/12/4-2023/12/10
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
-
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- 12/4 1038. 从二叉搜索树到更大和树
- 12/5 2477. 到达首都的最少油耗
- 12/6 2646. 最小化旅行的价格总和
- 12/7 1466. 重新规划路线
- 12/8 2008. 出租车的最大盈利
- 12/9 2048. 下一个更大的数值平衡数
- 12/10 70. 爬楼梯
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12/4 1038. 从二叉搜索树到更大和树
根据题目即 每个节点的值加上其右子树节点综合
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def bstToGst(root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
def dfs(node,fvalue):
if not node:
return fvalue
rv=dfs(node.right,fvalue)
node.val+=rv
lv=dfs(node.left,node.val)
return lv
dfs(root,0)
return root
12/5 2477. 到达首都的最少油耗
dfs 对于某一节点
为了省油 我们需要将其子路径上的节点汇聚到这个节点中 坐满车 再向终点前进
def minimumFuelCost(roads, seats):
"""
:type roads: List[List[int]]
:type seats: int
:rtype: int
"""
from collections import defaultdict
g = defaultdict(list)
for a,b in roads:
g[a].append(b)
g[b].append(a)
global ans
ans = 0
def dfs(cur,p):
global ans
s = 1
for nxt in g[cur]:
if nxt!=p:
cnt = dfs(nxt,cur)
s += cnt
ans +=(cnt+seats-1)//seats
return s
dfs(0,-1)
return ans
12/6 2646. 最小化旅行的价格总和
dfs 先统计所有路径中各个节点的贡献次数
ans[0] ans[1]表示该节点保持不变 和 减半的值
如果该节点减半 则相邻节点不能减半
def minimumTotalPrice(n, edges, price, trips):
"""
:type n: int
:type edges: List[List[int]]
:type price: List[int]
:type trips: List[List[int]]
:rtype: int
"""
ch = [[] for _ in range(n)]
for e in edges:
ch[e[0]].append(e[1])
ch[e[1]].append(e[0])
cnt = [0]*n
def dfs(node,p,end):
if node==end:
cnt[node]+=1
return True
for c in ch[node]:
if c==p:
continue
if dfs(c,node,end):
cnt[node]+=1
return True
return False
for x,y in trips:
dfs(x,-1,y)
def dp(node,p):
ans = [price[node]*cnt[node],price[node]*cnt[node]//2]
for c in ch[node]:
if c==p:
continue
[x,y] = dp(c,node)
ans[0],ans[1] = ans[0]+min(x,y),ans[1]+x
return ans
return min(dp(0,-1))
12/7 1466. 重新规划路线
因为路径为单向 所有点能到0
可以从0开始 考虑能够到达0的点
dfs
def minReorder(n, connections):
"""
:type n: int
:type connections: List[List[int]]
:rtype: int
"""
e = [[] for _ in range(n)]
for ed in connections:
e[ed[0]].append((ed[1],1))
e[ed[1]].append((ed[0],0))
def dfs(x,p):
ans = 0
for ed in e[x]:
if ed[0]==p:
continue
ans += ed[1]+dfs(ed[0],x)
return ans
return dfs(0,-1)
12/8 2008. 出租车的最大盈利
用哈希表m[end]记录终点为end的所有乘客信息
dpi记录到达第i个地点是的最大盈利
dp[i] = max(dp[i-1],dp[startj]+endj-startj+tipj)
def maxTaxiEarnings(n, rides):
"""
:type n: int
:type rides: List[List[int]]
:rtype: int
"""
m={}
dp = [0]*(n+1)
for r in rides:
if r[1] not in m:
m[r[1]]=[]
m[r[1]].append(r)
for i in range(1,n+1):
dp[i] = dp[i-1]
if i not in m:
continue
for r in m[i]:
dp[i] = max(dp[i],dp[r[0]]+r[1]-r[0]+r[2])
return dp[n]
12/9 2048. 下一个更大的数值平衡数
在范围内 平衡数是有限的
二分查找比n大的值
def nextBeautifulNumber( n):
"""
:type n: int
:rtype: int
"""
values = [
1, 22, 122, 212, 221, 333, 1333, 3133, 3313, 3331, 4444,
14444, 22333, 23233, 23323, 23332, 32233, 32323, 32332,
33223, 33232, 33322, 41444, 44144, 44414, 44441, 55555,
122333, 123233, 123323, 123332, 132233, 132323, 132332,
133223, 133232, 133322, 155555, 212333, 213233, 213323,
213332, 221333, 223133, 223313, 223331, 224444, 231233,
231323, 231332, 232133, 232313, 232331, 233123, 233132,
233213, 233231, 233312, 233321, 242444, 244244, 244424,
244442, 312233, 312323, 312332, 313223, 313232, 313322,
321233, 321323, 321332, 322133, 322313, 322331, 323123,
323132, 323213, 323231, 323312, 323321, 331223, 331232,
331322, 332123, 332132, 332213, 332231, 332312, 332321,
333122, 333212, 333221, 422444, 424244, 424424, 424442,
442244, 442424, 442442, 444224, 444242, 444422, 515555,
551555, 555155, 555515, 555551, 666666, 1224444]
l,r=0,len(values)-1
while l<r:
mid = (l+r)>>1
if values[mid]<=n:
l=mid+1
else:
r=mid
return values[l]
12/10 70. 爬楼梯
一次爬1或2 对于位置n来说 可以由n-1或n-2过来
所以记录前两次的方法 当前位置是前两次相加
def climbStairs(n):
"""
:type n: int
:rtype: int
"""
pre = 0
now = 1
for i in range(n):
next = now + pre
pre = now
now = next
return now