239. Sliding Window Maximum

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?

Similar:

76. Minimum Window Substring

155. Min Stack

159. Longest Substring with At Most Two Distinct Characters

265. Paint House II

Solution 1. Priority Queue
public class Solution {
class Pair implements Comparable<Pair>{
public int key;
public int idx;

public Pair(int k, int id) {
this.key = k;
this.idx = id;
}

@Override
public int compareTo(Pair p) {
return this.key < p.key ? 1 : (this.key == p.key ? 0 : -1); // reverse order, from Large to Small
}
}

public int[] maxSlidingWindow(int[] nums, int k) {
if (nums.length == 0 || k <= 0) return new int[0];

PriorityQueue<Pair> que = new PriorityQueue<Pair>();
int[] ret = new int[nums.length - k + 1];
int i = 0;

for (i = 0; i < nums.length; i++) {
que.add(new Pair(nums[i], i));
while (que.peek().idx < i - k + 1) {
que.poll();
}
if (i >= k-1) {
ret[i-k+1] = que.peek().key;
}
}

return ret;
}
}