LeetCode 每日一题 2025/3/10-2025/3/16

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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3/10 2269. 找到一个数字的 K 美丽值

依次判断

def divisorSubstrings(num, k):
    """
    :type num: int
    :type k: int
    :rtype: int
    """
    n=len(str(num))
    cur = num
    MOD = 10**k
    ans=0
    for i in range(n-k+1):
        v = cur%MOD
        if v!=0 and num%v==0:
            ans+=1
        cur//=10
    return ans

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3/11 2012. 数组美丽值求和

maxl[i]记录从0~i的最大值 minr[i]记录从i~n的最小值
对于nums[x] 如果maxl[x-1]<nums[x]<minr[x+1]则美丽值为2

def sumOfBeauties(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    n=len(nums)
    maxl=[0]*n
    minr=[0]*n
    maxl[0]=nums[0]
    minr[-1]=nums[-1]
    for i in range(1,n):
        maxl[i]=max(maxl[i-1],nums[i])
        minr[n-1-i]=min(minr[n-i],nums[n-1-i])
    ans = 0
    for i in range(1,n-1):
        if maxl[i-1]<nums[i] and nums[i]<minr[i+1]:
            ans+=2
        elif nums[i-1]<nums[i] and nums[i]<nums[i+1]:
            ans+=1
    return ans

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3/12 3305. 元音辅音字符串计数 I

恰好k个 可以看作至少k+1个 减去 至少k个
双指针
tag用来记录元音字母出现次数

def countOfSubstrings(word, k):
    """
    :type word: str
    :type k: int
    :rtype: int
    """
    def func(k):
        l=0
        ans = 0
        curk = 0
        tag = {}
        for w in word:
            if w in "aeiou":
                tag[w] = tag.get(w,0)+1
            else:
                curk+=1
            while len(tag)==5 and curk>=k:
                lw = word[l]
                if lw in "aeiou":
                    tag[lw]-=1
                    if tag[lw]==0:
                        del tag[lw]
                else:
                    curk-=1
                l+=1
            ans+=l
        return ans
    return func(k)-func(k+1)

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3/13 3306. 元音辅音字符串计数 II

恰好k个 可以看作至少k+1个 减去 至少k个
双指针
tag用来记录元音字母出现次数

def countOfSubstrings(word, k):
    """
    :type word: str
    :type k: int
    :rtype: int
    """
    def func(k):
        l=0
        ans = 0
        curk = 0
        tag = {}
        for w in word:
            if w in "aeiou":
                tag[w] = tag.get(w,0)+1
            else:
                curk+=1
            while len(tag)==5 and curk>=k:
                lw = word[l]
                if lw in "aeiou":
                    tag[lw]-=1
                    if tag[lw]==0:
                        del tag[lw]
                else:
                    curk-=1
                l+=1
            ans+=l
        return ans
    return func(k)-func(k+1)
                    

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3/14 3340. 检查平衡字符串

将奇偶位置的数值相减 判断最后差值是否为0

def isBalanced(num):
    """
    :type num: str
    :rtype: bool
    """
    cur=0
    n=len(num)
    for i in range(n):
        if i%2:
            cur+=int(num[i])
        else:
            cur-=int(num[i])
    return False if cur else True

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3/15 3110. 字符串的分数

按要求依次计算

def scoreOfString(s):
    """
    :type s: str
    :rtype: int
    """
    n=len(s)
    ans = 0
    for i in range(n-1):
        ans += abs(ord(s[i+1])-ord(s[i]))
    return ans

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3/16

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