From 4619998f5b7f4b2f16fc20f9649ac91a568a20d8 Mon Sep 17 00:00:00 2001
From: CyC2018 <zhengyc101@163.com>
Date: Wed, 28 Oct 2026 13:45:07 +0800
Subject: [PATCH] =?UTF-8?q?Update=2059.=20=E6=BB=91=E5=8A=A8=E7=AA=97?=
 =?UTF-8?q?=E5=8F=A3=E7=9A=84=E6=9C=80=E5=A4=A7=E5=80=BC.md?=
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题目链接改为力扣中的对应题目，题目描述和代码按该题的接口调整。
---
 notes/59. 滑动窗口的最大值.md | 27 +++++++++++++++------------
 1 file changed, 15 insertions(+), 12 deletions(-)

diff --git a/notes/59. 滑动窗口的最大值.md b/notes/59. 滑动窗口的最大值.md
index ed3f39d7..6958656e 100644
--- a/notes/59. 滑动窗口的最大值.md	
+++ b/notes/59. 滑动窗口的最大值.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/1624bc35a45c42c0bc17d17fa0cba788?tpId=13&tqId=11217&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/hua-dong-chuang-kou-de-zui-da-zhi-lcof/)
 
 ## 题目描述
 
@@ -19,18 +19,21 @@
 假设窗口的大小为 M，数组的长度为 N。在窗口向右移动时，需要先在堆中删除离开窗口的元素，并将新到达的元素添加到堆中，这两个操作的时间复杂度都为 log<sub>2</sub>M，因此算法的时间复杂度为 O(Nlog<sub>2</sub>M)，空间复杂度为 O(M)。
 
 ```java
-public ArrayList<Integer> maxInWindows(int[] num, int size) {
-    ArrayList<Integer> ret = new ArrayList<>();
-    if (size > num.length || size < 1)
-        return ret;
+public int[] maxAltitude(int[] heights, int limit) {
+    if (heights == null || heights.length == 0 || limit <= 0) {
+        return new int[0];
+    }
+    int[] ret = new int[heights.length - limit + 1];
     PriorityQueue<Integer> heap = new PriorityQueue<>((o1, o2) -> o2 - o1);  /* 大顶堆 */
-    for (int i = 0; i < size; i++)
-        heap.add(num[i]);
-    ret.add(heap.peek());
-    for (int i = 0, j = i + size; j < num.length; i++, j++) {            /* 维护一个大小为 size 的大顶堆 */
-        heap.remove(num[i]);
-        heap.add(num[j]);
-        ret.add(heap.peek());
+    for (int i = 0; i < limit; i++) {
+        heap.add(heights[i]);
+    }
+    int idx = 0;
+    ret[idx++] = heap.peek();
+    for (int i = 0, j = i + limit; j < heights.length; i++, j++) {            /* 维护一个大小为 limit 的大顶堆 */
+        heap.remove(heights[i]);
+        heap.add(heights[j]);
+        ret[idx++] = heap.peek();
     }
     return ret;
 }