LeetCode 每日一题 2023/2/27-2023/3/5

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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2/27 1144. 递减元素使数组呈锯齿状

两种情况 第一种奇数位的小 第二种偶数位的小
从头遍历

def movesToMakeZigzag(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    n = len(nums)
    ans1,ans2 = 0,0
    for i in range(n):
        tmp1,tmp2 = 0,0
        if i%2==0:
            if i>0 and nums[i]>=nums[i-1]:
                tmp1 = nums[i]-nums[i-1]+1
            if i<n-1 and nums[i]>=nums[i+1]:
                tmp2 = nums[i]-nums[i+1]+1
            ans1 += max(tmp1,tmp2)
        else:
            if nums[i]>=nums[i-1]:
                tmp1 = nums[i]-nums[i-1]+1
            if i<n-1 and nums[i]>=nums[i+1]:
                tmp2 = nums[i]-nums[i+1]+1
            ans2 += max(tmp1,tmp2)
    return min(ans1,ans2)

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2/28 2363. 合并相似的物品

依次遍历 vlist记录所有出现过的价值
集合统计每个价值的重量和

def mergeSimilarItems(items1, items2):
    """
    :type items1: List[List[int]]
    :type items2: List[List[int]]
    :rtype: List[List[int]]
    """
    m = {}
    vlist = set()
    for v,w in items1:
        m[v] = m.get(v,0)+w
        vlist.add(v)
    for v,w in items2:
        m[v] = m.get(v,0)+w
        vlist.add(v)
    ans = []
    for v in sorted(list(vlist)):
        ans.append([v,m[v]])
    return ans
        

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3/1 2373. 矩阵中的局部最大值

每行每列依次寻找相连三格最大值

def largestLocal(grid):
    """
    :type grid: List[List[int]]
    :rtype: List[List[int]]
    """
    n = len(grid)
    tmp = [[0]*n for _ in range(n)]
    for i in range(n):
        for j in range(1,n-1):
            tmp[i][j] = max(grid[i][j-1],grid[i][j],grid[i][j+1])
   
    ans = [[0]*n for _ in range(n)]
    for i in range(n):
        for j in range(1,n-1):
            ans[j][i] = max(tmp[j][i],tmp[j-1][i],tmp[j+1][i])

    return [ans[i][1:n-1] for i in range(1,n-1)]

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3/2 面试题 05.02. 二进制数转字符串

小数点后一位一位相加 直至超过32位

def printBin(num):
    """
    :type num: float
    :rtype: str
    """
    curr = 1.0
    ans = "0."
    while len(ans)<32:
        curr = curr/2
        if num>=curr:
            num-=curr
            ans+="1"
        else:
            ans+="0"
        if num==0:
            return ans
    return "ERROR"

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3/3 1487. 保证文件名唯一

map存储所有出现过的名字 并记录其当前编号

def getFolderNames(names):
    """
    :type names: List[str]
    :rtype: List[str]
    """
    ans = []
    m = {}
    for name in names:
        if name not in m:
            ans.append(name)
            m[name]=1
        else:
            ind = m[name]
            while name+'('+str(ind)+')' in m:
                ind+=1
            cur = name+'('+str(ind)+')'
            ans.append(cur)
            m[name] = ind+1
            m[cur] = 1
    return ans

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3/4 982. 按位与为零的三元组

先统计两数相与能够得到的结果cnt
再依次统计每个数在cnt中有多少能与其相与为0

def countTriplets(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    cnt = [0]*(1<<16)
    for i in nums:
        for j in nums:
            cnt[i&j] +=1
    ans = 0
    for num in nums:
        target = num ^ 0xffff
        s = target
        while True:
            ans += cnt[s]
            s = (s-1)&target
            if s==target:
                break
    return ans

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3/5 1599. 经营摩天轮的最大利润

如果4个人的钱小于运行的钱 则必定亏本
依次遍历每个时间点的游客 wait记录当前等待游客数量
ans记录最大利润时的经营时间 cur记录当前利润 maxv记录最大利润
当没有后续游客时 继续考虑等待的游客 每次上4人使得利润最大化

def minOperationsMaxProfit(customers, boardingCost, runningCost):
    """
    :type customers: List[int]
    :type boardingCost: int
    :type runningCost: int
    :rtype: int
    """
    if 4*boardingCost<runningCost:
        return -1
    wait = 0
    ans = -1
    cur = -999
    maxv = -999
    num = 0
    for cus in customers:
        wait += cus
        tmp = min(4,wait)
        wait -=tmp
        cur += tmp*boardingCost-runningCost
        num +=1
        if cur>maxv:
            maxv = cur
            ans = num
    while wait>0:
        tmp = min(4,wait)
        wait -=tmp
        cur += tmp*boardingCost-runningCost
        num +=1
        if cur>maxv:
            maxv = cur
            ans = num
    return ans

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