LeetCode 每日一题 2024/10/28-2024/11/3

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

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10/28 685. 冗余连接 II

check用来检查是否满足
1.入度不能有大于1的
2.不能存在圈
先将所有点的父节点取出，如果存在有两个父节点 那么说明这个节点需要取出一条边 遍历后check是否正确
若正确 将结果暂存ret
判断ret,
若ret内有多个 则取出位置最靠后的作为结果
若ret==0:
所有节点入度都正常 说明存在圈 从后往前去除一个点 check是否正常 若正常则答案为这个点

def findRedundantDirectedConnection(edges):""":type edges: List[List[int]]:rtype: List[int]"""def find(p,x):while p[x] != x:p[x] = p[p[x]]x = p[x]return p[x]# 判断给定的图是不是有圈def cycle(graph):p = {}for a, b in graph:p.setdefault(a, a)p.setdefault(b, b)pa, pb = find(p, a), find(p, b)print([a,b],p,pa,pb)if pa == pb: return(True)else:p[pa] = p[pb]dic={}ans=[]ret =[]def check(l):#入度不能大于1endpoint = [y for x,y in l]if len(endpoint)!=len(set(endpoint)):return Falsedef find(p,x):while p[x] != x:p[x] = p[p[x]]x = p[x]return p[x]# 判断给定的图是不是有圈def cycle(graph):p = {}for a, b in graph:p.setdefault(a, a)p.setdefault(b, b)pa, pb = find(p, a), find(p, b)if pa == pb: return(True)else:p[pa] = p[pb]if cycle(l):return Falsereturn Truefor v in edges:top = v[0]end = v[1]dic.setdefault(end,[])dic[end].append(top)for k in dic.keys():if len(dic[k])>1:for v in dic[k]:ori = edges[:]tmp=[v,k]ori.pop(ori.index(tmp))if check(ori):ret.append(tmp)if len(ret)==0:for i in range(len(edges)-1,-1,-1):ori = edges[:]ori.pop(i)if check(ori):ans = edges[i]breakelse:pos = -1for i in ret:if edges.index(i)>pos:pos = edges.index(i)ans = i    return ans

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10/29 3211. 生成不含相邻零的二进制字符串

所有长度为2的子字符串至少有一个1 说明不存在连续的两个0
将二进制按位取反为x 判断不存在连续两个1
如果x&(x>>1)=0 说明x没有连续的两个1

def validStrings(n):""":type n: int:rtype: List[str]"""ans = []mask = (1<<n)-1for i in range(1<<n):t = mask^iif (t>>1 & t)==0:ans.append(f"{i:0{n}b}")return ans

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10/30 3216. 交换后字典序最小的字符串

从前往后寻找
找到首次出现满足前面数字比后面大的就交换

def getSmallestString(s):""":type s: str:rtype: str"""l=list(s)n=len(s)for i in range(n-1):if int(l[i])%2==int(l[i+1])%2 and l[i]>l[i+1]:l[i],l[i+1]=l[i+1],l[i]breakreturn ''.join(l)

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10/31 3165. 不包含相邻元素的子序列的最大和

将数组分为两段 每一段的首尾都有不选 或 可选两种情况
一共就有四种情况
00:首不选 尾不选
01:首不选 尾可选
10:首可选 尾不选
11:首可选 尾可选
将某数组x分为前后a,b两段 分别讨论四种情况
f(x)为x数组不相邻子序列最大和
f00(x) = max(f01(a)+f00(b),f00(a)+f10(b))
f01(x) = max(f00(a)+f11(b),f01(a)+f01(b))
f10(x) = max(f10(a)+f10(b),f11(a)+f00(b))
f11(x) = max(f10(a)+f11(b),f11(a)+f01(b))
用线段树来实现单点的修改 每个节点维护对应区间f(X)

def maximumSumSubsequence(nums, queries):""":type nums: List[int]:type queries: List[List[int]]:rtype: int"""MOD=10**9+7n=len(nums)tr = [[0]*4 for _ in range(2<<n.bit_length())]def maxv(a,b):return b if b>a else adef merg(k):a,b = tr[k*2],tr[k*2+1]tr[k][0] = max(a[0]+b[2],a[1]+b[0])tr[k][1] = max(a[0]+b[3],a[1]+b[1])tr[k][2] = max(a[2]+b[2],a[3]+b[0])tr[k][3] = max(a[2]+b[3],a[3]+b[1])def build(k,l,r):if l==r:tr[k][3]=maxv(nums[l],0)returnm=(l+r)//2build(k*2,l,m)build(k*2+1,m+1,r)merg(k)def change(k,l,r,i,val):if l==r:tr[k][3]=max(val,0)returnm=(l+r)//2if i<=m:change(k*2,l,m,i,val)else:change(k*2+1,m+1,r,i,val)merg(k)build(1,0,n-1)ans = 0for i,val in queries:change(1,0,n-1,i,val)ans = (ans+tr[1][3])%MODreturn ans

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11/1 3259. 超级饮料的最大强化能量

动态规划 f[i][0/1]代表第i步喝A/B获得的最大值

def maxEnergyBoost(energyDrinkA, energyDrinkB):""":type energyDrinkA: List[int]:type energyDrinkB: List[int]:rtype: int"""n=len(energyDrinkA)f = [[0,0] for _ in range(n+1)]for i in range(1,n+1):f[i][0] = f[i-1][0]+energyDrinkA[i-1]f[i][1] = f[i-1][1]+energyDrinkB[i-1]if i>1:f[i][0] = max(f[i][0],f[i-2][1]+energyDrinkA[i-1])f[i][1] = max(f[i][1],f[i-2][0]+energyDrinkB[i-1])return max(f[n][0],f[n][1])

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11/2 3226. 使两个整数相等的位更改次数

按位判断

def minChanges(n, k):""":type n: int:type k: int:rtype: int"""ans = 0while n>0 or k>0:if (n&1)==0 and (k&1)==1:return -1if (n&1)==1 and (k&1)==0:ans+=1n>>=1k>>=1return ans

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11/3638. 大礼包

过滤掉不需要的套餐 保留有用套餐
dfs 遍历选取每一种套餐的情况
算出不买套餐的价格base
如果买套餐实惠则替换价格

def shoppingOffers(price, special, needs):""":type price: List[int]:type special: List[List[int]]:type needs: List[int]:rtype: int"""n = len(price)fsp = []for sp in special:if sum(sp[i]*price[i] for i in range(n))>sp[n]:fsp.append(sp)def dfs(curneed):base = sum(curneed[i]*price[i] for i in range(n))for cursp in fsp:p = cursp[n]netneed = []for i in range(n):if cursp[i]>curneed[i]:breaknetneed.append(curneed[i]-cursp[i])if len(netneed)==n:base = min(base,dfs(netneed)+p)return basereturn dfs(needs)

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