LeetCode 题解（143）: Sliding Window Maximum

题目：

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 k numbers 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?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered.

题解：

用Heap可以实现O((n-k)lgk)复杂度。要求Linear Time的话只能借助Deque。Deque的队首始终保持最大元素。然后根据一定条件进Queue即可。

C++版：

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> result;
        if(nums.size() == 0 || k == 0)
            return result;
            
        deque<int> buffer;

        for(int i = 0; i < k; i++) {
            bufferModifier(buffer, nums[i]);
        }
        
        result.push_back(buffer.front());
        
        for(int i = k; i < nums.size(); i++) {
            if(buffer.front() == nums[i-k])
                buffer.pop_front();
            bufferModifier(buffer, nums[i]);
            result.push_back(buffer.front());
        }
        
        return result;
    }
    
    void bufferModifier(deque<int>& buffer, int n) {
        if(buffer.size() == 0)
            buffer.push_back(n);
        else if(buffer.front() < n) {
            buffer.clear();
            buffer.push_back(n);
        } else {
            while(buffer.back() < n)
                buffer.pop_back();
            buffer.push_back(n);
        }
    }
};

Java版：

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums.length == 0)
            return new int[0];
            
        Deque<Integer> buffer = new LinkedList<>();
        for(int i = 0; i < k; i++) {
            bufferModifier(buffer, nums[i]);
        }
        
        int[] result = new int[nums.length - k + 1];
        result[0] = buffer.getFirst();
        
        for(int i = k; i < nums.length; i++) {
            if(buffer.getFirst() == nums[i-k])
                buffer.pollFirst();
            bufferModifier(buffer, nums[i]);
            result[i-k+1] = buffer.getFirst();
        }
        return result;
    }
    
    public void bufferModifier(Deque<Integer> buffer, int n) {
        if(buffer.isEmpty()) {
            buffer.addFirst(n);
        } else if(n > buffer.getFirst()) {
            buffer.clear();
            buffer.addFirst(n);
        } else {
            while(buffer.getLast() < n)
                buffer.pollLast();
            buffer.addLast(n);
        }
    }
}

Python版：

class Solution:
    # @param {integer[]} nums
    # @param {integer} k
    # @return {integer[]}
    def maxSlidingWindow(self, nums, k):
        if len(nums) == 0:
            return []

        result = []
        buffer = []
        for i in range(k):
            self.bufferModifier(buffer, nums[i])

        result.append(buffer[0])

        for i in range(k, len(nums)):
            if buffer[0] == nums[i-k]:
                buffer = buffer[1:]
            self.bufferModifier(buffer, nums[i])
            result.append(buffer[0])

        return result

    def bufferModifier(self, buffer, n):
        if len(buffer) == 0:
            buffer.append(n)
        elif buffer[0] < n:
            del buffer[:]
            buffer.append(n)
        else:
            while buffer[len(buffer)-1] < n:
                buffer.pop()
            buffer.append(n)