LeetCode 每日一题 2023/11/27-2023/12/3

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

      • 11/27 907. 子数组的最小值之和
      • 11/28 1670. 设计前中后队列
      • 11/29 2336. 无限集中的最小数字
      • 11/30 1657. 确定两个字符串是否接近
      • 12/1 2661. 找出叠涂元素
      • 12/2 1094. 拼车
      • 12/3 1423. 可获得的最大点数


11/27 907. 子数组的最小值之和

对于某位置i arr[i]对答案的贡献 需要找到min(b)=arr[i]的数组
在i位置左侧有连续l个数大于arr[i] ,i位置右侧有连续r个大于arr[i]
那么子数组一共有m=(l+1)(r+1)个 min(b) = arr[i]
所以i位置的数贡献了m次 m
arr[i]
找左侧大于等于自己的个数
右侧同样的方法 为防止重复计算 右侧严格大于自己

def sumSubarrayMins(arr):
    """
    :type arr: List[int]
    :rtype: int
    """
    MOD = 10**9+7
    ans = 0
    n = len(arr)
    stack = []
    l = [0]*n
    r = [0]*n
    for i,v in enumerate(arr):
        while stack and v<=arr[stack[-1]]:
            stack.pop()
        if stack:
            l[i] = i - stack[-1]
        else:
            l[i] = i+1
        stack.append(i)
    stack = []
    for i in range(n-1,-1,-1):
        while stack and arr[i]<arr[stack[-1]]:
            stack.pop()
        if stack:
            r[i] = stack[-1]-i
        else:
            r[i] = n-i
        stack.append(i)
    for i in range(n):
        ans = (ans+l[i]*r[i]*arr[i])%MOD
    return ans



11/28 1670. 设计前中后队列

将队列分为前后两部分处理

class FrontMiddleBackQueue(object):

    def __init__(self):
        self.num = 0
        self.l = []
        self.r = []


    def pushFront(self, val):
        """
        :type val: int
        :rtype: None
        """
        self.num +=1
        self.l = [val]+self.l
        if len(self.l)>len(self.r):
            self.r = [self.l[-1]]+self.r
            self.l = self.l[:-1]


    def pushMiddle(self, val):
        """
        :type val: int
        :rtype: None
        """
        self.num += 1
        if len(self.l)==len(self.r):
            self.r = [val]+self.r
        else:
            self.l.append(val)


    def pushBack(self, val):
        """
        :type val: int
        :rtype: None
        """
        self.num += 1
        self.r.append(val)
        if len(self.r)>len(self.l)+1:
            self.l.append(self.r[0])
            self.r = self.r[1:]


    def popFront(self):
        """
        :rtype: int
        """
        v = -1
        if self.num==0:
            return v
        if len(self.l)>0:
            v = self.l[0]
            self.l = self.l[1:]
        else:
            v = self.r[0]
            self.r = self.r[1:]
        self.num -=1
        if len(self.r)>len(self.l)+1:
            self.l.append(self.r[0])
            self.r = self.r[1:]
        return v


    def popMiddle(self):
        """
        :rtype: int
        """
        v = -1
        if self.num==0:
            return v
        if len(self.l)==len(self.r):
            v = self.l[-1]
            self.l = self.l[:-1]
        else:
            v = self.r[0]
            self.r = self.r[1:]
        self.num-=1
        return v


    def popBack(self):
        """
        :rtype: int
        """
        v = -1
        if self.num==0:
            return v
        v=self.r[-1]
        self.num-=1
        self.r = self.r[:-1]
        if len(self.l)>len(self.r):
            self.r = [self.l[-1]]+self.r
            self.l = self.l[:-1]
        return v



11/29 2336. 无限集中的最小数字

记录已经丢弃的数集合 s
记录当前最小数 cur

class SmallestInfiniteSet(object):

    def __init__(self):
        self.s = set()
        self.cur = 1


    def popSmallest(self):
        """
        :rtype: int
        """
        v = self.cur
        self.s.add(v)
        self.cur += 1
        while self.cur in self.s:
            self.cur+=1
        return v


    def addBack(self, num):
        """
        :type num: int
        :rtype: None
        """
        if num in self.s:
            self.s.remove(num)
        if num < self.cur:
            self.cur = num




11/30 1657. 确定两个字符串是否接近

统计所有出现次数 只要次数相同 两个字符串即接近

def closeStrings(word1, word2):
    """
    :type word1: str
    :type word2: str
    :rtype: bool
    """
    from collections import Counter
    if len(word1)!=len(word2):
        return False
    return set(word1)==set(word2) and Counter(Counter(word1).values())==Counter(Counter(word2).values())



12/1 2661. 找出叠涂元素

遍历矩阵记录每个数的位置
遍历arr 记录每行每列当前未被涂色的个数


def firstCompleteIndex(arr, mat):
    """
    :type arr: List[int]
    :type mat: List[List[int]]
    :rtype: int
    """
    m,n=len(mat),len(mat[0])
    mem = {}
    for i in range(m):
        for j in range(n):
            mem[mat[i][j]]=(i,j)
            
    row = [n]*m
    col = [m]*n
    
    for i,v in enumerate(arr):
        a,b = mem[v]
        row[a]-=1
        col[b]-=1
        if row[a]==0 or col[b]==0:
            return i
        


12/2 1094. 拼车

最小堆储存需要下车的信息(t,num) 在位置t需要下num个人
当前为f 将所有小于等于f的下车人数都下车

def carPooling(trips, capacity):
    """
    :type trips: List[List[int]]
    :type capacity: int
    :rtype: bool
    """
    import heapq
    trips.sort(key = lambda x:(x[1],x[2]))
    cur = 0
    m = []
    for num,f,t in trips:
        while m and m[0][0]<=f: 
            loc,v = heapq.heappop(m)
            cur -= v
        cur += num
        if cur>capacity:
            return False
        heapq.heappush(m, (t,num))
    return True



12/3 1423. 可获得的最大点数

取前后k个数 剩余中间n-k个数
滑动窗口判断n-k个数和最小

def maxScore(cardPoints, k):
    """
    :type cardPoints: List[int]
    :type k: int
    :rtype: int
    """
    total = sum(cardPoints)
    l = len(cardPoints)-k
    cur  = sum(cardPoints[:l])
    s = cur
    for i in range(len(cardPoints)-l):
        print(cur,cardPoints[i],cardPoints[i+l])
        cur = cur-cardPoints[i]+cardPoints[i+l]
        s = min(s,cur)
    return total - s



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