一刷697

Author: diguage

README.adoc

@@ -4904,13 +4904,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0696-count-binary-substrings.adoc[题解]
 //|Easy
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/degree-of-an-array/[697. Degree of an Array^]
-//|{source_base_url}/_0697_DegreeOfAnArray.java[Java]
-//|{doc_base_url}/0697-degree-of-an-array.adoc[题解]
-//|Easy
-//|
+
+|{counter:codes}
+|{leetcode_base_url}/degree-of-an-array/[697. Degree of an Array^]
+|{source_base_url}/_0697_DegreeOfAnArray.java[Java]
+|{doc_base_url}/0697-degree-of-an-array.adoc[题解]
+|Easy
+|
 
 |{counter:codes}
 |{leetcode_base_url}/partition-to-k-equal-sum-subsets/[698. Partition to K Equal Sum Subsets^]

docs/0697-degree-of-an-array.adoc

@@ -1,46 +1,66 @@
 [#0697-degree-of-an-array]
-= 697. Degree of an Array
+= 697. 数组的度
 
-{leetcode}/problems/degree-of-an-array/[LeetCode - Degree of an Array^]
+https://leetcode.cn/problems/degree-of-an-array/[LeetCode - 697. 数组的度 ^]
 
-Given a non-empty array of non-negative integers `nums`, the *degree* of this array is defined as the maximum frequency of any one of its elements.
-Your task is to find the smallest possible length of a (contiguous) subarray of `nums`, that has the same degree as `nums`.
+给定一个非空且只包含非负数的整数数组 `nums`，数组的 *度* 的定义是指数组里任一元素出现频数的最大值。
 
-*Example 1:*
+你的任务是在 `nums` 中找到与 `nums` 拥有相同大小的度的最短连续子数组，返回其长度。
 
+*示例 1：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [1, 2, 2, 3, 1]
-*Output:* 2
-*Explanation:* 
-The input array has a degree of 2 because both elements 1 and 2 appear twice.
-Of the subarrays that have the same degree:
+....
+输入：nums = [1,2,2,3,1]
+输出：2
+解释：
+输入数组的度是 2 ，因为元素 1 和 2 的出现频数最大，均为 2 。
+连续子数组里面拥有相同度的有如下所示：
 [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
-The shortest length is 2. So return 2.
-----
-
+最短连续子数组 [2, 2] 的长度为 2 ，所以返回 2 。
+....
 
+*示例 2：*
 
-*Example 2:*
+....
+输入：nums = [1,2,2,3,1,4,2]
+输出：6
+解释：
+数组的度是 3 ，因为元素 2 重复出现 3 次。
+所以 [2,2,3,1,4,2] 是最短子数组，因此返回 6 。
+....
 
+*提示：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [1,2,2,3,1,4,2]
-*Output:* 6
-----
-
+* `nums.length` 在 `1` 到 `50,000` 范围内。
+* `nums[i]` 是一个在 `0` 到 `49,999` 范围内的整数。
 
-*Note:*
-. `nums.length` will be between 1 and 50,000.</li>
-. `nums[i]` will be an integer between 0 and 49,999.</li>
 
+== 思路分析
 
 
 [[src-0697]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0697_DegreeOfAnArray.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0697_DegreeOfAnArray_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
+
 

docs/index.adoc

@@ -1476,7 +1476,7 @@ include::0695-max-area-of-island.adoc[leveloffset=+1]
 
 // include::0696-count-binary-substrings.adoc[leveloffset=+1]
 
-// include::0697-degree-of-an-array.adoc[leveloffset=+1]
+include::0697-degree-of-an-array.adoc[leveloffset=+1]
 
 include::0698-partition-to-k-equal-sum-subsets.adoc[leveloffset=+1]
 

logbook/202503.adoc

@@ -544,10 +544,15 @@ endif::[]
 |✅ 贪心算法。最高位尽可能跟后面的最大数字进行交换。
 
 |{counter:codes2503}
-|{leetcode_base_url}/two-sum-ii-input-array-is-sorted/[167. Two Sum II - Input array is sorted^]
+|{leetcode_base_url}/two-sum-ii-input-array-is-sorted/[167. 两数之和 II - 输入有序数组^]
 |{doc_base_url}/0167-two-sum-ii-input-array-is-sorted.adoc[题解]
 |✅ 双指针
 
+|{counter:codes2503}
+|{leetcode_base_url}/degree-of-an-array/[697. 数组的度^]
+|{doc_base_url}/0697-degree-of-an-array.adoc[题解]
+|✅
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0697_DegreeOfAnArray.java

@@ -0,0 +1,49 @@
+package com.diguage.algo.leetcode;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+public class _0697_DegreeOfAnArray {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-29 09:15:49
+   */
+  public int findShortestSubArray(int[] nums) {
+    Map<Integer, List<Integer>> numToIndex = new HashMap<>();
+    Map<Integer, Integer> indexToDegre = new HashMap<>();
+    int max = 0;
+
+    for (int i = 0; i < nums.length; i++) {
+      int num = nums[i];
+      List<Integer> index = numToIndex.get(num);
+      if (index == null) {
+        index = new ArrayList<>();
+        index.add(i);
+        index.add(i);
+        numToIndex.put(num, index);
+      } else {
+        index.set(1, i);
+      }
+      int count = indexToDegre.getOrDefault(num, 0) + 1;
+      max = Math.max(max, count);
+      indexToDegre.put(num, count);
+    }
+    int minLength = Integer.MAX_VALUE;
+    for (Map.Entry<Integer, Integer> entry : indexToDegre.entrySet()) {
+      if (max == entry.getValue()) {
+        List<Integer> index = numToIndex.get(entry.getKey());
+        minLength = Math.min(minLength, index.get(1) - index.get(0));
+      }
+    }
+    return minLength + 1;
+  }
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0697_DegreeOfAnArray()
+      .findShortestSubArray(new int[]{2, 1, 1, 2, 1, 3, 3, 3, 1, 3, 1, 3, 2});
+  }
+}