[LeetCode] 239.滑动窗口的最大值

方法一：

这道题跟左神上课讲的题一样，可以使用一个双端队列来对存数据。

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int[] list=new int[nums.length-k+1];
        LinkedList<Integer> deque=new LinkedList<>();
        int index=0;
        for(int i=0;i<nums.length;i++){
            while(!deque.isEmpty()&&nums[deque.peekLast()]<=nums[i]){
                deque.pollLast();
            }
            deque.add(i);
            if(i-deque.peekFirst()==k){
                deque.pollFirst();
            }
            if(i>=k-1){
                list[index++]=nums[deque.peekFirst()];
            }
        }
        return list;
    }
}

 方法二：动态规划（官方给出的解答）

class Solution {
  public int[] maxSlidingWindow(int[] nums, int k) {
    int n = nums.length;
    if (n * k == 0) return new int[0];
    if (k == 1) return nums;

    int [] left = new int[n];
    left[0] = nums[0];
    int [] right = new int[n];
    right[n - 1] = nums[n - 1];
    for (int i = 1; i < n; i++) {
      // from left to right
      if (i % k == 0) left[i] = nums[i];  // block_start
      else left[i] = Math.max(left[i - 1], nums[i]);

      // from right to left
      int j = n - i - 1;
      if ((j + 1) % k == 0) right[j] = nums[j];  // block_end
      else right[j] = Math.max(right[j + 1], nums[j]);
    }

    int [] output = new int[n - k + 1];
    for (int i = 0; i < n - k + 1; i++)
      output[i] = Math.max(left[i + k - 1], right[i]);

    return output;
  }
}