//给定一个数组 nums，有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k 个数字。滑动窗口每次只向右移动一位。 
//
//
// 返回滑动窗口中的最大值。 
//
// 
//
// 进阶： 
//
// 你能在线性时间复杂度内解决此题吗？ 
//
// 
//
// 示例: 
//
// 输入: nums = [1,3,-1,-3,5,3,6,7], 和 k = 3
//输出: [3,3,5,5,6,7] 
//解释: 
//
//  滑动窗口的位置                最大值
//---------------               -----
//[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 
//
// 
//
// 提示： 
//
// 
// 1 <= nums.length <= 10^5 
// -10^4 <= nums[i] <= 10^4 
// 1 <= k <= nums.length 
// 
// Related Topics 堆 Sliding Window

package ;

//Java：滑动窗口最大值
public class P239SlidingWindowMaximum {
    public static void main(String[] args) {
        Solution solution = new P239SlidingWindowMaximum().new Solution();
        // TO TEST
        int[] nums = {1, 3, -1, -3, 5, 3, 6, 7};
        int k = 3;
        int[] res = solution.maxSlidingWindow(nums, k);
        for (int i = 0; i < res.length; i++) {
            System.out.println(res[i]);
        }
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public int[] maxSlidingWindow(int[] nums, int k) {
            if (k == 0) {
                return null;
            }
            int[] res = new int[nums.length - k + 1];
            int resIndex = 0;
            for (int i = 0,count = 0; count <= nums.length-k; i++,count++) {
                res[resIndex++] = maxValue(nums, i, k);
            }

            return res;
        }

        int maxValue(int[] nums, int start, int len) {
            int max = nums[start];
            for (int i = start; i < len+start; i++) {
                max = max < nums[i] ? nums[i] : max;
            }
            return max;
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}