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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4/291146. 快照数组

快照时无需记录整个数组
只要记录与前一份快照之间的修改
使用history[x]记录x位置元素的修改记录(snapid,value)
查询一个ind位置某个id的数据时
只要在history[ind]中二分找到快照编号小于id的最后一条修改记录值即可

class SnapshotArray(object):def __init__(self, length):""":type length: int"""from collections import defaultdictself.curid = 0self.history = defaultdict(list)def set(self, index, val):""":type index: int:type val: int:rtype: None"""self.history[index].append((self.curid,val))def snap(self):""":rtype: int"""self.curid +=1return self.curid-1def get(self, index, snap_id):""":type index: int:type snap_id: int:rtype: int"""import bisectj = bisect.bisect_left(self.history[index], (snap_id+1,))-1return self.history[index][j][1] if j>=0 else 0

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4/30 2798. 满足目标工作时长的员工数目

依次遍历判断是否达到要求

def numberOfEmployeesWhoMetTarget(hours, target):""":type hours: List[int]:type target: int:rtype: int"""ans = 0for h in hours:if h>=target:ans+=1return ans

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5/1 2462. 雇佣 K 位工人的总代价

建立两个最小堆 如果左右可选择个数加上被选择k个包含了所有的数值
那么直接返回最小的k个数之和即可

def totalCost(costs, k, candidates):""":type costs: List[int]:type k: int:type candidates: int:rtype: int"""import heapqn = len(costs)if candidates*2+k>n:costs.sort()return sum(costs[:k])pre = costs[:candidates]suf = costs[-candidates:]heapq.heapify(pre)heapq.heapify(suf)ans = 0l,r=candidates,n-candidates-1for _ in range(k):if pre[0]<=suf[0]:ans += heapq.heapreplace(pre, costs[l])l+=1else:ans += heapq.heapreplace(suf, costs[r])r-=1return ans

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5/2 857. 雇佣 K 名工人的最低成本

r=wage/quality
单位工作质量的工资从小到大排序
假设k个工人质量和为sumQ 以r为基准发工资 那么总额为sumQ*r
最大堆维护k个quality 获取更小的sumQ

def mincostToHireWorkers(quality, wage, k):""":type quality: List[int]:type wage: List[int]:type k: int:rtype: float"""import heapqqw = sorted(zip(quality,wage),key = lambda x:1.0*x[1]/x[0])h = [-q for q,_ in qw[:k]]heapq.heapify(h)totalq = -sum(h)ans = totalq*1.0*qw[k-1][1]/qw[k-1][0]for q,w in qw[k:]:if q<-h[0]:totalq += heapq.heapreplace(h,-q)+qans = min(ans,totalq*1.0*w/q)return ans

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5/3 1491. 去掉最低工资和最高工资后的工资平均值

求总和 最大值 最小值

def average(salary):""":type salary: List[int]:rtype: float"""s = sum(salary)minv = min(salary)maxv = max(salary)n = len(salary)return (s-minv-maxv)/(n-2)

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5/4 1235. 规划兼职工作

将工作按结束时间从小到大排列
dp[i]记录截止前i个工作能获得的最大收益
第i-1份工作可以选择做 或者 不做
选择做则需要找到结束时间小于等于开始时间的位置

def jobScheduling(startTime, endTime, profit):""":type startTime: List[int]:type endTime: List[int]:type profit: List[int]:rtype: int"""n = len(startTime)job = sorted(zip(startTime,endTime,profit),key = lambda x:x[1])dp = [0]*(n+1)li = sorted(endTime)def find(r,v):l = 0while l<=r:mid = (l+r)>>1if li[mid]> v:r = mid-1else:l = mid+1return lfor i in range(1,n+1):k = find(i,job[i-1][0])dp[i] = max(dp[i-1],dp[k]+job[i-1][2])return dp[n]

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5/5 1652. 拆炸弹

三种情况分别判断

def decrypt(code, k):""":type code: List[int]:type k: int:rtype: List[int]"""n = len(code)ans = [0]*nif k>0:cur = sum(code[:k])for i in range(n):cur = cur-code[i]+code[(k+i)%n]ans[i] = curelif k<0:cur = sum(code[n+k:])for i in range(n):ans[i] = curcur = cur+code[i]-code[(n+k+i)%n]return ans

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