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L0239_SlidingWindowMaximum.java
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110 lines (98 loc) · 3.39 KB
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import java.util.Deque;
import java.util.LinkedList;
/**
* https://leetcode.cn/problems/sliding-window-maximum/
*
* 给你一个整数数组 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。
*
* 示例 1:
* 输入:nums = [1,3,-1,-3,5,3,6,7], 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
*
* 示例 2:
* 输入:nums = [1], k = 1
* 输出:[1]
*
* 提示:
* - 1 <= nums.length <= 10⁵
* - -10⁴ <= nums[i] <= 10⁴
* - 1 <= k <= nums.length
*/
public class L0239_SlidingWindowMaximum {
/**
* 使用双端队列实现滑动窗口最大值
* 队列中存储的是数组下标,并且保持队列中的元素单调递减
*/
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0 || k <= 0) {
return new int[0];
}
int n = nums.length;
int[] result = new int[n - k + 1];
Deque<Integer> deque = new LinkedList<>();
// 处理第一个窗口
for (int i = 0; i < k; i++) {
// 保持队列单调递减
while (!deque.isEmpty() && nums[i] >= nums[deque.peekLast()]) {
deque.pollLast();
}
deque.offerLast(i);
}
// 记录第一个窗口的最大值
result[0] = nums[deque.peekFirst()];
// 处理后续窗口
for (int i = k; i < n; i++) {
// 移除不在当前窗口范围内的元素
while (!deque.isEmpty() && deque.peekFirst() <= i - k) {
deque.pollFirst();
}
// 保持队列单调递减
while (!deque.isEmpty() && nums[i] >= nums[deque.peekLast()]) {
deque.pollLast();
}
deque.offerLast(i);
// 记录当前窗口的最大值
result[i - k + 1] = nums[deque.peekFirst()];
}
return result;
}
public static void main(String[] args) {
L0239_SlidingWindowMaximum solution = new L0239_SlidingWindowMaximum();
// 测试用例 1
int[] nums1 = {1, 3, -1, -3, 5, 3, 6, 7};
int k1 = 3;
int[] result1 = solution.maxSlidingWindow(nums1, k1);
System.out.print("Test Case 1: [");
for (int i = 0; i < result1.length; i++) {
System.out.print(result1[i]);
if (i < result1.length - 1) {
System.out.print(", ");
}
}
System.out.println("]");
// 测试用例 2
int[] nums2 = {1};
int k2 = 1;
int[] result2 = solution.maxSlidingWindow(nums2, k2);
System.out.print("Test Case 2: [");
for (int i = 0; i < result2.length; i++) {
System.out.print(result2[i]);
if (i < result2.length - 1) {
System.out.print(", ");
}
}
System.out.println("]");
}
}