Leetcode 239 Sliding Window Maximum

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Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the knumbers in the window. Each time the sliding window moves right by one position.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3
 1 [3  -1  -3] 5  3  6  7       3
 1  3 [-1  -3  5] 3  6  7       5
 1  3  -1 [-3  5  3] 6  7       5
 1  3  -1  -3 [5  3  6] 7       6
 1  3  -1  -3  5 [3  6  7]      7

Therefore, return the max sliding window as [3,3,5,5,6,7].

Note: 
You may assume k is always valid, ie: 1 ≤ k ≤ input array‘s size for non-empty array.

Follow up:
Could you solve it in linear time?

O(n^2),deque maxlen, 当deque超过设定长度自动删除之前的元素。

class Solution:
    # @param {integer[]} nums
    # @param {integer} k
    # @return {integer[]}
    def maxSlidingWindow(self, nums, k):
        d = collections.deque(maxlen=k)
        ans = []
        for i in range(len(nums)):
            d.append(nums[i])
            ans.append(max(d))
        return ans[k-1:]

O(n) 类似于Kadane算法思想，如果之前的元素都小与即将遇到的元素，则删光之前所有的元素，反之则加入当前元素并且检测是否框架长度超出。

class Solution:
    # @param {integer[]} nums
    # @param {integer} k
    # @return {integer[]}
    def maxSlidingWindow(self, nums, k):
        d = collections.deque()
        ans = []
        for i in range(len(nums)):
            while d and nums[d[-1]] < nums[i]:
                d.pop()
            d.append(i)
            if d[0] == i-k:
                d.popleft()
            ans.append(nums[d[0]])
        return ans[k-1:]

Leetcode 239 Sliding Window Maximum

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原文地址：http://www.cnblogs.com/lilixu/p/4657202.html