25期代码随想录算法训练营第十三天 | 栈与队列 part 2

239. 滑动窗口最大值

链接

窗口 — 维持一个单调递增队列

为什么要使用队列？

• 在窗口移动的时候，方便把不属于窗口的最大值剔除。（当窗口移动之后）

class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        q = deque() # store index
        res = []
        l = r = 0

        while r < len(nums):
            # push
            # make sure that it's a mono increasing queue
            while q and nums[q[-1]] < nums[r]:
                q.pop()
            q.append(r)

            # pop element 
            # since it's not in the window anymore
            if l > q[0]:
                q.popleft()

            # append the max num in the window to res
            # only when a window is formed can we move the l pointer
            if (r + 1) >= k:
                res.append(nums[q[0]])
                l += 1
            r += 1
        return res
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347.前 K 个高频元素

链接

方法一

Counter得到频率表，反转频率表，再进行heapify。
heappop前k个元素，即为前K个高频元素。

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        _map = Counter(nums)
        _map = [(-v, i) for i, v in _map.items()]
        heapify(_map)
        res = []
        for i in range(k):
            want = heappop(_map)
            res.append(want[1])
        return res
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方法二

Bucket Sort

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        _map = Counter(nums)
        freq = [[] for i in range(len(nums) + 1)]
        for key, val in _map.items():
            freq[val].append(key)

        res = []
        for i in range(len(freq) - 1, 0, -1):
            while freq[i] and k > 0:
                res.append(freq[i].pop())
                k -= 1
        
            if k == 0:
                break
        
        return res
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