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
- 7/24 771. 宝石与石头
- 7/25 2208. 将数组和减半的最少操作次数
- 7/26 2569. 更新数组后处理求和查询
- 7/27 2500. 删除每行中的最大值
- 7/28 2050. 并行课程 III
- 7/29 141. 环形链表
- 7/30
7/24 771. 宝石与石头
将宝石类型放入set中
一次判断石头中宝石个数
def numJewelsInStones(jewels, stones):""":type jewels: str:type stones: str:rtype: int"""j = set(jewels)ans = 0for s in stones:if s in j:ans+=1return ans7/25 2208. 将数组和减半的最少操作次数
大顶堆记录当前最大值 每次取最大值减半
def halveArray(nums):""":type nums: List[int]:rtype: int"""import heapqtarget = sum(nums)*1.0/2l = []for num in nums:heapq.heappush(l,-num)cur = 0ans = 0while True:v = -heapq.heappop(l)cur +=v*1.0/2ans+=1if cur>=target:breakheapq.heappush(l,-v*1.0/2)return ans7/26 2569. 更新数组后处理求和查询
操作1是将nums1中[l,r]的数反转
操作2是将nums2的和加上p*sum(nums1)
操作3是求nums2的和
只要考虑nums1的变化即可
线段树维护nums1区间
l,r左右端点 s区间和 lazy懒标记
class Node:def __init__(self):self.l=self.r=0self.s = 0self.lazy = 0
class SegmentTree:def __init__(self,nums):self.nums = numsn = len(nums)self.tr = [Node() for _ in range(n<<2)]self.build(1,1,n)def build(self,u,l,r):self.tr[u].l,self.tr[u].r=l,rif l==r:self.tr[u].s = self.nums[l-1]returnmid = (l+r)>>1self.build(u<<1,l,mid)self.build(u<<1|1,mid+1,r)self.pushup(u)def modify(self,u,l,r):if self.tr[u].l>=l and self.tr[u].r<=r:self.tr[u].lazy^=1self.tr[u].s = self.tr[u].r-self.tr[u].l+1-self.tr[u].sreturnself.pushdown(u)mid = (self.tr[u].l+self.tr[u].r)>>1if l<=mid:self.modify(u<<1,l,r)if r>mid:self.modify(u<<1|1,l,r)self.pushup(u)def query(self,u,l,r):if self.tr[u].l>=l and self.tr[u].r<=r:return self.tr[u].sself.pushdown(u)mid = (self.tr[u].l+self.tr[u].r)>>1ans = 0if l<=mid:ans +=self.query(u<<1,l,r)if r>mid:ans +=self.query(u<<1|1,l,r)return ansdef pushup(self,u):self.tr[u].s = self.tr[u<<1].s+self.tr[u<<1|1].sdef pushdown(self,u):if self.tr[u].lazy:mid = (self.tr[u].l+self.tr[u].r)>>1self.tr[u<<1].s = mid-self.tr[u].l+1-self.tr[u<<1].sself.tr[u<<1].lazy^=1self.tr[u<<1|1].s = self.tr[u].r-mid-self.tr[u<<1|1].sself.tr[u<<1|1].lazy^=1self.tr[u].lazy^=1def handleQuery(nums1, nums2, queries):""":type nums1: List[int]:type nums2: List[int]:type queries: List[List[int]]:rtype: List[int]"""tr = SegmentTree(nums1)s = sum(nums2)ans = []for op,l,r in queries:if op==1:tr.modify(1,l+1,r+1)elif op==2:s+=l*tr.query(1,1,len(nums1))else:ans.append(s)return ans7/27 2500. 删除每行中的最大值
每一行从小到大排序
再取每一列的最大值
def deleteGreatestValue(grid):""":type grid: List[List[int]]:rtype: int"""cur =[]for l in grid:cur.append(sorted(l))return sum([max([cur[i][j] for i in range(len(cur))]) for j in range(len(grid[0]))])7/28 2050. 并行课程 III
m[i]记录课程i为多少课程的前置课程
ind[i]记录课程i有几个前置课程
t[i]记录课程i的最早完成时间
def minimumTime(n, relations, time):""":type n: int:type relations: List[List[int]]:type time: List[int]:rtype: int"""from collections import defaultdictm = defaultdict(list)ind=[0]*nfor pre,nxt in relations:m[pre-1].append(nxt-1)ind[nxt-1]+=1l = []t=[0]*nans = 0for i in range(n):if ind[i]==0:l.append(i)t[i]=time[i]ans = max(ans,t[i])while l:tmp = []for i in l:for j in m[i]:t[j] = max(t[j],t[i]+time[j])ans = max(ans,t[j])ind[j]-=1if ind[j]==0:tmp.append(j)l=tmpreturn ans7/29 141. 环形链表
快慢指针
class ListNode(object):def __init__(self, x):self.val = xself.next = None
def hasCycle(head):""":type head: ListNode:rtype: bool"""if head==None or head.next==None:return Falseslow = headfast = head.nextwhile(slow!=fast):if fast==None or fast.next==None:return Falseslow=slow.nextfast=fast.next.nextreturn True7/30
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