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[LeetCode] 239. Sliding Window Maximum

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. Return the max sliding window.

Example:

Input: nums =          [1,3,-1,-3,5,3,6,7]                , and k = 3
Output:          [3,3,5,5,6,7] 
Explanation: 
                
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      

方法一:

[LeetCode] 239. Sliding Window Maximum
public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums==null||nums.length<k||k==0){
            return new int[0];
        }
        if(k==1){
            return nums;
        }
        int len=nums.length;
        int []ans=new int[len-k+1];
        for(int i=0;i<k;i++){
            ans[0]=Math.max(ans[0],nums[i]);
        }
        for(int i=k;i<len;i++){
            if(ans[i-k]!=nums[i-k]){
                ans[i-k+1]=Math.max(ans[i-k],nums[i]);
            }else{
                for(int j=i-k+1;j<=i;j++){
                    ans[i-k+1]=Math.max(ans[i-k+1],nums[j]);
                }
            }
        }
        return ans;
    }
           

方法二:优先级队列: 来源 https://www.cnblogs.com/lightwindy/p/8661524.html

[LeetCode] 239. Sliding Window Maximum
public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums==null||nums.length<k||k==0){
            return new int[0];
        }
        if(k==1){
            return nums;
        }
        PriorityQueue<Integer> queue=new PriorityQueue<Integer>(Collections.reverseOrder());
        int len=nums.length;
        int []ans=new int[len-k+1];
        for(int i=0;i<len;i++){
            if(i>=k){
                queue.remove(nums[i-k]);
            }
            queue.offer(nums[i]);
            if(i-k+1>=0){
                ans[i-k+1]=queue.peek();
            }
        }
        return ans;
}
           

优先级队列相关详解可查看:https://blog.csdn.net/liuwenyou/article/details/100535592