5946. 句子中的最多单词数

这是我写的：执行用时: 32 ms，内存消耗: 15.2 MB

class Solution:
    def maxSubsequence(self, nums: List[int], k: int) -> List[int]:
        ans = 1
        for sentence in sentences:
            sen = sentence.split(' ')
            ans = max(ans, len(sen))
        return ans

这是大佬写的：执行用时: 32 ms，内存消耗: 15.1 MB

class Solution:
    def maxSubsequence(self, nums: List[int], k: int) -> List[int]:
        return max(len(sentence.split()) for sentence in sentences)

我用了额外空间，属实没必要~

5947. 从给定原材料中找到所有可以做出的菜

这题真的能看出来算法的博大精深，因为我的用时和大佬的差距太太太大了。 这是我写的：执行用时: 9904 ms，内存消耗: 16.6 MB

class Solution:
    def findAllRecipes(self, recipes: List[str], ingredients: List[List[str]], supplies: List[str]) -> List[str]:
        lenr, leni, lens = len(recipes), len(ingredients), len(supplies)
        isinsup = [False for i in range(lenr)]
        flag = True
        while flag:
            flag = False
            for i in range(lenr):
                if not isinsup[i] and all(ing in supplies for ing in ingredients[i]):
                    flag = True
                    supplies.append(recipes[i])
                    isinsup[i] = True
        return supplies[lens:]

这是大佬写的：执行用时: 176 ms，内存消耗: 16.7 MB，双100%

class Solution:
    def findAllRecipes(self, recipes: List[str], ingredients: List[List[str]], supplies: List[str]) -> List[str]:
        # 建图，并记录入度indegree
        graph = collections.defaultdict(list)
        indeg = collections.defaultdict(int)
        for u, vs in zip(recipes, ingredients):
            for v in vs:
                graph[v].append(u)
                indeg[u] += 1
        
        # 将原材料作为迭代的起始点【题目已满足：给定的原材料在图中的入度均为0】
        deque = collections.deque(supplies)

        # 从原材料开始迭代，并更新indegree
        while deque:
            cur = deque.popleft()
            for nxt in graph[cur]:    # 遍历当前 cur 节点的邻居/子代节点
                indeg[nxt] -= 1
                if indeg[nxt] == 0:   # indegree=0则加入队列
                    deque.append(nxt)
            
        res = [rec for rec in recipes if indeg[rec] == 0] 
        return res

5948. 判断一个括号字符串是否有效

大佬写的：执行用时: 248 ms，内存消耗: 15.5 MB

class Solution:
    def canBeValid(self, s: str, locked: str) -> bool:
        if len(s)%2 == 1:
            return False
        # 正序遍历：未匹配的左括号 ( 的最大数目
        cnt = 0
        for ch, b in zip(s, locked):
            if ch == '(' and b == '1':
                cnt += 1
            elif ch == ')' and b == '1':
                cnt -= 1
            elif b == '0':
                cnt += 1   
            if cnt < 0:
                return False
        # 逆序遍历：未匹配的右括号 ) 的最大数目
        cnt = 0
        for ch, b in zip(s[::-1], locked[::-1]):
            if ch == ')' and b == '1':
                cnt += 1
            elif ch == '(' and b == '1':
                cnt -= 1
            elif b == '0':
                cnt += 1
            if cnt < 0:
                return False
        return True

5949. 一个区间内所有数乘积的缩写

大佬写的：执行用时: 3388 ms，内存消耗: 15 MB

class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        if right - left < 30 :
            to_ret = 1
            for n in range(left, right+1) :
                to_ret *= n
            n0 = 0
            while to_ret % 10 == 0 :
                n0 += 1
                to_ret = to_ret // 10
            to_ret = str(to_ret)
            if len(to_ret) <= 10 :
                return to_ret + 'e' + str(n0)
            else :
                return to_ret[:5] + '...' + to_ret[-5:] + 'e' + str(n0)
                
        n0 = 0
        to_reth = 1
        to_rett = 1
        for n in range(left, right+1) :
            to_reth = to_reth * n
            while to_reth >= 10**5 :
                to_reth = to_reth / 10
            
            to_rett = to_rett * n
            while to_rett % 10 == 0 :
                to_rett = to_rett // 10
                n0 += 1
            to_rett = to_rett % (10**32)
        return str(int(to_reth)) + '...' + str(to_rett)[-5:] + 'e' + str(n0)