leetcode 第 309 场周赛

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检查相同字母间的距离

标记每次出现的字母所出现的位置，进行判断即可

class Solution:
    def checkDistances(self, s: str, distance: List[int]) -> bool:
        vis = [0] * 26
        for step, ss in enumerate(s):
            ss = ord(ss) - ord('a')
            if vis[ss] and step - vis[ss] != distance[ss]:
                return False
            vis[ss] = step + 1
        return True

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恰好移动 k 步到达某一位置的方法数目

第一眼看上去有点复杂，但想想用记忆化搜索即可解决
dfs中注意要对结果取余

class Solution:
    def numberOfWays(self, startPos: int, endPos: int, k: int) -> int:
        Mo = 1e9+7
        @cache
        def dfs(n, k):
            if abs(n) > k: 
                 return 0
            if k == 0: 
                 return n == 0
            return (dfs(n-1, k-1) + dfs(n+1, k-1)) % Mo
        return int(dfs(endPos-startPos, k))

python 可以直接用装饰器实现记忆化还真挺方便的

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最长优雅子数组

又看错题目了，原来说的是子数组，还以为是子序列就直接放弃了
子数组的话直接使用滚动窗口解决

class Solution:
    def longestNiceSubarray(self, nums: List[int]) -> int:
        left,mask,res = 0,0,0
        for right, num in enumerate(nums):      
            while mask & num > 0:       
                mask ^= nums[left]     
                left += 1                       
            mask ^= num              
            res = max(res, right-left+1)
        return res

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会议室 III

题目还挺长的，不过仔细阅读一下就理解了
感觉以前做过类似的题目，那时候用一个堆来维护
这里使用两个堆，一个维护空闲会议室，一个维护使用中的会议室

class Solution:
    def mostBooked(self, n: int, meetings: List[List[int]]) -> int:
        vis = [0] * n
        rest, use = list(range(n)), []
        meetings=sorted(meetings,key=lambda x: x[0])
        for start, end in meetings:
            while use and use[0][0] <= start:
                heappush(rest, heappop(use)[1])
            if rest:
                pos = heappop(rest)
            else:
                temp, pos = heappop(use) 
                end += temp - start
            vis[pos] += 1
            heappush(use, (end, pos))
        res = 0
        for pos, r in enumerate(vis):
            if r > vis[res]:
                res = pos
        return res

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总结

orz