一刷480

Author: diguage

README.adoc

@@ -3385,14 +3385,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0479-largest-palindrome-product.adoc[题解]
 //|Hard
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/sliding-window-median/[480. Sliding Window Median^]
-//|{source_base_url}/_0480_SlidingWindowMedian.java[Java]
-//|{doc_base_url}/0480-sliding-window-median.adoc[题解]
-//|Hard
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/sliding-window-median/[480. Sliding Window Median^]
+|{source_base_url}/_0480_SlidingWindowMedian.java[Java]
+|{doc_base_url}/0480-sliding-window-median.adoc[题解]
+|Hard
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/magical-string/[481. Magical String^]
 //|{source_base_url}/_0481_MagicalString.java[Java]

docs/0436-find-right-interval.adoc

@@ -58,11 +58,27 @@ For [2,3], the interval [3,4] has minimum-"right" start point.
 
 *NOTE:* input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
 
-
+== 思路分析
 
 [[src-0436]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0436_FindRightInterval.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0005_LongestPalindromicSubstring_2.java[tag=answer]
+// ----
+// --
+====
 

docs/0480-sliding-window-median.adoc

@@ -39,11 +39,32 @@ You may assume `k` is always valid, ie: `k` is always smaller than input array's
 
 Answers within `10^-5` of the actual value will be accepted as correct.
 
-
+== 思路分析
 
 [[src-0480]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0480_SlidingWindowMedian.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0005_LongestPalindromicSubstring_2.java[tag=answer]
+// ----
+// --
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/sliding-window-median/solutions/588643/hua-dong-chuang-kou-zhong-wei-shu-by-lee-7ai6/[480. 滑动窗口中位数 - 官方题解^]
+. https://leetcode.cn/problems/sliding-window-median/solutions/589561/xiang-jie-po-su-jie-fa-you-xian-dui-lie-mo397/[480. 滑动窗口中位数 - 详解「朴素解法」&「优先队列 」^]
 

docs/index.adoc

@@ -1034,7 +1034,7 @@ include::0462-minimum-moves-to-equal-array-elements-ii.adoc[leveloffset=+1]
 
 // include::0479-largest-palindrome-product.adoc[leveloffset=+1]
 
-// include::0480-sliding-window-median.adoc[leveloffset=+1]
+include::0480-sliding-window-median.adoc[leveloffset=+1]
 
 // include::0481-magical-string.adoc[leveloffset=+1]
 

logbook/202406.adoc

@@ -429,6 +429,11 @@
 |{doc_base_url}/0295-find-median-from-data-stream.adoc[题解]
 |双堆思路太妙了！
 
+|{counter:codes}
+|{leetcode_base_url}/sliding-window-median/[480. Sliding Window Median^]
+|{doc_base_url}/0480-sliding-window-median.adoc[题解]
+|双端队列+滑动窗口
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

src/main/java/com/diguage/algo/leetcode/_0295_FindMedianFromDataStream.java

@@ -10,30 +10,30 @@ public class _0295_FindMedianFromDataStream {
    * @since 2024-08-29 20:43:20
    */
   class MedianFinder {
-    Queue<Integer> big;
-    Queue<Integer> small;
+    Queue<Integer> topSmall;
+    Queue<Integer> topLarge;
 
     public MedianFinder() {
       // 小顶堆，顶部是最小的。保存较大的一半
-      big = new PriorityQueue<>();
+      topSmall = new PriorityQueue<>();
       // 大顶堆，顶部是最大的。保存较小的一半
-      small = new PriorityQueue<>((a, b) -> b - a);
+      topLarge = new PriorityQueue<>((a, b) -> b - a);
     }
 
     public void addNum(int num) {
-      if (big.size() != small.size()) {
-        // 倒腾一下，实际 small 中的元素增加了
-        big.add(num);
-        small.add(big.poll());
+      if (topSmall.size() != topLarge.size()) {
+        // 倒腾一下，实际 topLarge 中的元素增加了
+        topSmall.add(num);
+        topLarge.add(topSmall.poll());
       } else {
-        // 倒腾一下，实际 big 中的元素增加了
-        small.add(num);
-        big.add(small.poll());
+        // 倒腾一下，实际 topSmall 中的元素增加了
+        topLarge.add(num);
+        topSmall.add(topLarge.poll());
       }
     }
 
     public double findMedian() {
-      return big.size() != small.size() ? big.peek() : (big.peek() + small.peek()) / 2.0;
+      return topSmall.size() != topLarge.size() ? topSmall.peek() : (topSmall.peek() + topLarge.peek()) / 2.0;
     }
   }
   // end::answer[]

src/main/java/com/diguage/algo/leetcode/_0436_FindRightInterval.java

@@ -0,0 +1,10 @@
+package com.diguage.algo.leetcode;
+
+public class _0436_FindRightInterval {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-08-29 20:49:07
+   */
+  // end::answer[]
+}

src/main/java/com/diguage/algo/leetcode/_0480_SlidingWindowMedian.java

@@ -0,0 +1,60 @@
+package com.diguage.algo.leetcode;
+
+import java.util.PriorityQueue;
+import java.util.Queue;
+
+public class _0480_SlidingWindowMedian {
+  // tag::answer[]
+
+  /**
+   * 答案是正确的，超大数组测试时，超时了。
+   *
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-08-30 11:45:50
+   */
+  public double[] medianSlidingWindow(int[] nums, int k) {
+    // 如果窗口是奇数，则 topSmall 多一个
+    Queue<Integer> topSmall = new PriorityQueue<>((a, b) -> Integer.compare(a, b));
+    // 使用 Integer.compare，防止两个最小负数相减时溢出
+    Queue<Integer> topLarge = new PriorityQueue<>((a, b) -> Integer.compare(b, a));
+    double[] result = new double[nums.length - k + 1];
+    for (int i = 0; i < k; i++) {
+      topSmall.add(nums[i]);
+    }
+    for (int i = 0; i < k / 2; i++) {
+      topLarge.add(topSmall.poll());
+    }
+    // 先相处再相加，防止溢出
+    result[0] = k % 2 == 1 ? 1.0 * topSmall.peek()
+      : topSmall.peek() / 2.0 + topLarge.peek() / 2.0;
+    for (int i = k; i < nums.length; i++) {
+      int num = nums[i];
+      if (topSmall.peek() <= num) {
+        topSmall.add(num);
+      } else {
+        topLarge.add(num);
+      }
+
+      int delNum = nums[i - k];
+      if (topSmall.peek() <= delNum) {
+        topSmall.remove(delNum);
+      } else {
+        topLarge.remove(delNum);
+      }
+      // 添加，删除，再平衡双堆，这样才能保证堆的平衡性
+      // topSmall 应该始终大于等于 topLarge
+      while (topLarge.size() > topSmall.size()) {
+        topSmall.add(topLarge.poll());
+      }
+      // topSmall 与 topLarge 的差值不能大于 1
+      while (topSmall.size() - topLarge.size() > 1) {
+        topLarge.add(topSmall.poll());
+      }
+
+      result[i - k + 1] = k % 2 == 1 ? 1.0 * topSmall.peek()
+        : topSmall.peek() / 2.0 + topLarge.peek() / 2.0;
+    }
+    return result;
+  }
+  // end::answer[]
+}