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
- 2/17 1287. 有序数组中出现次数超过25%的元素
- 2/18 2080. 区间内查询数字的频率
- 2/19 624. 数组列表中的最大距离
- 2/20 2595. 奇偶位数
- 2/21 2209. 用地毯覆盖后的最少白色砖块
- 2/22 2506. 统计相似字符串对的数目
- 2/23 1206. 设计跳表
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 bisect
n=len(arr)
num=n//4+1
for 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
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0left,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 0
if li[0]>right or li[-1]<left:
return 0
ansl,ansr=0,0
if left<=li[0]:
ansl = 0
else:
l,r = 0,len(li)
while l<=r:
mid = l+(r-l)//2
if li[mid]==left:
l = mid
break
elif li[mid]>left:
r = mid-1
else:
l = mid+1
ansl = l
if right>=li[-1]:
ansr = len(li)-1
else:
l,r = 0,len(li)
while l<=r:
mid = l+(r-l)//2
if li[mid]==right:
r = mid
break
elif li[mid]>right:
r = mid-1
else:
l = mid+1
ansr = r
if ansr<ansl:
return 0
return ansr-ansl+1
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 = 0
for 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
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 = 0
even,odd=0,0
while n:
if n%2==1:
if cur%2:
odd+=1
else:
even+=1
cur+=1
n//=2
return [even,odd]
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]=0
for 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]
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 = 0
for c in w:
cur |= (1<<(ord(c)-ord('a')))
tag[cur] = tag.get(cur,0)+1
ans = 0
for v in tag.values():
ans += v*(v-1)//2
return ans
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 = 1
while lv<MAX_LEVAL and random.random()<PRO:
lv+=1
return lv
class Node:
__slots__ = ['val','forward']
def __init__(self,val,maxlev=MAX_LEVAL):
self.val = val
self.forward = [None]*maxlev
class Skiplist(object):
def __init__(self):
self.head = Node(-1)
self.level = 0
def search(self, target):
"""
:type target: int
:rtype: bool
"""
cur = self.head
for 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 True
return False
def add(self, num):
"""
:type num: int
:rtype: None
"""
new = [self.head]*MAX_LEVAL
cur = self.head
for i in range(self.level-1,-1,-1):
while cur.forward[i] and cur.forward[i].val<num:
cur = cur.forward[i]
new[i] = cur
lv = 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] = newnode
def erase(self, num):
"""
:type num: int
:rtype: bool
"""
new = [self.head]*MAX_LEVAL
cur = self.head
for i in range(self.level-1,-1,-1):
while cur.forward[i] and cur.forward[i].val<num:
cur = cur.forward[i]
new[i] = cur
cur = cur.forward[0]
if not cur or cur.val!=num:
return False
for i in range(self.level):
if new[i].forward[i]!=cur:
break
new[i].forward[i] = cur.forward[i]
while self.level>1 and not self.head.forward[self.level-1]:
self.level-=1
return True