LeetCode 每日一题 2023/10/23-2023/10/29

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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10/23 2678. 老人的数目

取第12-13位数值

def countSeniors(details):""":type details: List[str]:rtype: int"""ans = 0for d in details:age = int(d[11:13])if age>60:ans +=1return ans

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10/24 1155. 掷骰子等于目标和的方法数

递归 mem记录已有过的情况

def numRollsToTarget(n, k, target):""":type n: int:type k: int:type target: int:rtype: int"""if n>target or target>n*k:return 0MOD = 10**9+7mem = {(1,x):1 for x in range(1,k+1)}def check(n,target):if target<=0:return 0if (n,target) in mem:return mem[(n,target)]ans = 0if n==0:ans = target==0else:for i in range(1,k+1):ans += check(n-1,target-i)mem[(n,target)] = ansreturn ans%MODreturn check(n,target)

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10/25 2698. 求一个整数的惩罚数

遍历每个数 判断是否满足条件

def punishmentNumber(n):""":type n: int:rtype: int"""def check(s,i,x):m = len(s)if i>=m:return x==0y = 0for j in range(i,m):y = y*10+int(s[j])if y>x:breakif check(s,j+1,x-y):return Truereturn Falseans = 0for i in range(1,n+1):if check(str(i*i),0,i):ans+=i*ireturn ans

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10/26 2520. 统计能整除数字的位数

按每一位数字判断

def countDigits(num):""":type num: int:rtype: int"""ans = 0n = numwhile n:v = n%10if v>0 and num%v==0:ans +=1n//=10return ans

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10/27 1465. 切割后面积最大的蛋糕

将切口位置从小到大排序 计算两个切口间的距离
寻找最大距离 相乘

def maxArea(h, w, horizontalCuts, verticalCuts):""":type h: int:type w: int:type horizontalCuts: List[int]:type verticalCuts: List[int]:rtype: int"""horizontalCuts.sort()verticalCuts.sort()if len(horizontalCuts)>0:maxh = max(horizontalCuts[0],h-horizontalCuts[-1])else:maxh = hif len(verticalCuts)>0:maxv = max(verticalCuts[0],w-verticalCuts[-1])else:maxv = wfor i in range(1,len(horizontalCuts)):maxh = max(maxh,horizontalCuts[i]-horizontalCuts[i-1])for i in range(1,len(verticalCuts)):maxv = max(maxv,verticalCuts[i]-verticalCuts[i-1])return (maxv*maxh)%(10**9+7)

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10/28 2558. 从数量最多的堆取走礼物

大顶堆 每次取最大的礼物 进行处理

def pickGifts(gifts, k):""":type gifts: List[int]:type k: int:rtype: int"""import heapq,mathans = sum(gifts)l = [-x for x in gifts]heapq.heapify(l)for i in range(k):print(l)v = -heapq.heappop(l)num = int(math.sqrt(v))ans -= (v-num)heapq.heappush(l, -num)return ans

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10/29

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