艺纵网站建设建立网站平台需要多少钱

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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2/17 1287. 有序数组中出现次数超过25%的元素

存在这个元素至少出现num次 那么这个元素必定会出现在位置0，num,2*num…处
找到这些位置 寻找该位置的值出现次数

def findSpecialInteger(arr):""":type arr: List[int]:rtype: int"""import bisectn=len(arr)num=n//4+1for i in range(0,n,num):if bisect.bisect_right(arr, arr[i])-bisect.bisect_left(arr, arr[i])>=num:return arr[i]return -1

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2/18 2080. 区间内查询数字的频率

使用一个map m[value]存放数值x出现的位置
求取m[value]中有多少个位置在[left,right]内
分别求0_left,0right内位置个数 相减即可

from collections import defaultdict
class RangeFreqQuery(object):def __init__(self, arr):""":type arr: List[int]"""self.m = defaultdict(list)for i,v in enumerate(arr):self.m[v].append(i)def query(self, left, right, value):""":type left: int:type right: int:type value: int:rtype: int"""li = self.m[value]if len(li)==0:return 0if li[0]>right or li[-1]<left:return 0ansl,ansr=0,0if left<=li[0]:ansl = 0else:l,r = 0,len(li)while l<=r:mid = l+(r-l)//2if li[mid]==left:l = midbreakelif li[mid]>left:r = mid-1else:l = mid+1ansl = lif right>=li[-1]:ansr = len(li)-1else:l,r = 0,len(li)while l<=r:mid = l+(r-l)//2if li[mid]==right:r = midbreakelif li[mid]>right:r = mid-1else:l = mid+1ansr = rif ansr<ansl:return 0return ansr-ansl+1

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2/19 624. 数组列表中的最大距离

遍历每个数组 记录最小值 最大值

def maxDistance(arrays):""":type arrays: List[List[int]]:rtype: int"""mi,ma=arrays[0][0],arrays[0][-1]ans = 0for i in range(1,len(arrays)):curmi,curma=arrays[i][0],arrays[i][-1]ans=max(ans,abs(ma-curmi),abs(curma-mi))mi=min(mi,curmi)ma=max(ma,curma)return ans

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2/20 2595. 奇偶位数

依次判断当前最低位是否为一 判断好后每次除以2

def evenOddBit(n):""":type n: int:rtype: List[int]"""cur = 0even,odd=0,0while n:if n%2==1:if cur%2:odd+=1else:even+=1cur+=1n//=2return [even,odd]

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2/21 2209. 用地毯覆盖后的最少白色砖块

dp[i][j]表示前i个砖用了j个地毯后最少的白色砖块

def minimumWhiteTiles(floor, numCarpets, carpetLen):""":type floor: str:type numCarpets: int:type carpetLen: int:rtype: int"""n=len(floor)dp = [[float("inf")] * (numCarpets+1) for _ in range(n+1)]for j in range(numCarpets+1):dp[0][j]=0for i in range(1,n+1):dp[i][0] = dp[i-1][0]+(1 if floor[i-1]=='1' else 0)for i in range(1,n+1):for j in range(1,numCarpets+1):dp[i][j]=dp[i-1][j]+(1 if floor[i-1]=='1' else 0)dp[i][j]=min(dp[i][j],dp[max(0,i-carpetLen)][j-1])return dp[n][numCarpets]

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2/22 2506. 统计相似字符串对的数目

使用26位0，1 记录组成字符串的字母出现情况
计算相同字母的字符串个数能够组成的对数

def similarPairs(words):""":type words: List[str]:rtype: int"""tag={}for w in words:cur = 0for c in w:cur |= (1<<(ord(c)-ord('a')))tag[cur] = tag.get(cur,0)+1ans = 0for v in tag.values():ans += v*(v-1)//2return ans

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2/23 1206. 设计跳表

参考 https://leetcode.cn/problems/design-skiplist/solution/she-ji-tiao-biao-by-leetcode-solution-e8yh/
将每个数字看作一个节点 节点最多有MAX_LEVAL层
如果对于某一层 该节点存在 则该位置指向下一个存在的节点
random_level 随机决定该节点层数

import random
MAX_LEVAL = 32
PRO = 0.25
def random_level():lv = 1while lv<MAX_LEVAL and random.random()<PRO:lv+=1return lvclass Node:__slots__ = ['val','forward']def __init__(self,val,maxlev=MAX_LEVAL):self.val = valself.forward = [None]*maxlevclass Skiplist(object):def __init__(self):self.head = Node(-1)self.level = 0def search(self, target):""":type target: int:rtype: bool"""cur = self.headfor i in range(self.level-1,-1,-1):while cur.forward[i] and cur.forward[i].val<target:cur = cur.forward[i]cur = cur.forward[0]if cur and cur.val==target:return Truereturn Falsedef add(self, num):""":type num: int:rtype: None"""new = [self.head]*MAX_LEVALcur = self.headfor i in range(self.level-1,-1,-1):while cur.forward[i] and cur.forward[i].val<num:cur = cur.forward[i]new[i] = curlv = random_level()self.level = max(self.level,lv)newnode = Node(num,lv)for i in range(lv):newnode.forward[i] = new[i].forward[i]new[i].forward[i] = newnodedef erase(self, num):""":type num: int:rtype: bool"""new = [self.head]*MAX_LEVALcur = self.headfor i in range(self.level-1,-1,-1):while cur.forward[i] and cur.forward[i].val<num:cur = cur.forward[i]new[i] = curcur = cur.forward[0]if not cur or cur.val!=num:return Falsefor i in range(self.level):if new[i].forward[i]!=cur:breaknew[i].forward[i] = cur.forward[i]while self.level>1 and not self.head.forward[self.level-1]:self.level-=1return True

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