13天【代码随想录算法训练营34期】 第五章 栈与队列part03(● 239. 滑动窗口最大值 ● 347.前 K 个高频元素)

239. 滑动窗口最大值
单调队列:单调递减,一个queue,最大值在queue口,队列中只维护有可能为最大值的数字
比如说1,3,2,4;当sliding window已经到3时,就可以把1 pop出去了,因为有了3,1不可能为最大值,同理到4的时候,3、2都可以pop出去

class MyQueue:
    def __init__(self):
        self.queue = deque()
    def pop(self, value):
        if self.queue and value == self.queue[0]:
            self.queue.popleft()
    def push(self, value):
        while self.queue and value > self.queue[-1]:
            self.queue.pop()
        self.queue.append(value)
    def front(self):
        return self.queue[0]

class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        q = MyQueue()
        result = []
        for i in range(k):
            q.push(nums[i])
        result.append(q.front())

        for i in range(k, len(nums)):
            q.pop(nums[i-k])
            q.push(nums[i])
            result.append(q.front())

        return result

347.前 K 个高频元素
先把数字和频率存进一个dictionary {key数字:value出现的次数}
大顶堆:root为最大数的tree
小顶堆:root为最小数的tree
如果找前k个最大的数,应该使用小顶堆,因为一般tree pop的都是root,用小顶堆的话,正好pop的就是最小值

import heapq
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        map_ = {}
        priority_q = []
        for i in range(len(nums)):
            if nums[i] in map_:
                map_[nums[i]] += 1
            else:
                map_[nums[i]] = 0
        for key, freq in map_.items():
            heapq.heappush(priority_q, (freq, key))
            if len(priority_q) > k:
                heapq.heappop(priority_q)
        
        result = [0] * k
        for i in range(k-1, -1, -1):
            result[i] = heapq.heappop(priority_q)[1]
        return result
posted @ 2024-04-01 11:07  MiraMira  阅读(24)  评论(0)    收藏  举报