一刷3115

Author: diguage

README.adoc

@@ -21831,12 +21831,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|Easy
 //|
 
-//|{counter:codes}
-//|{leetcode_base_url}/maximum-prime-difference/[3115. Maximum Prime Difference^]
-//|{source_base_url}/_3115_MaximumPrimeDifference.java[Java]
-//|{doc_base_url}/3115-maximum-prime-difference.adoc[题解]
-//|Medium
-//|
+|{counter:codes}
+|{leetcode_base_url}/maximum-prime-difference/[3115. Maximum Prime Difference^]
+|{source_base_url}/_3115_MaximumPrimeDifference.java[Java]
+|{doc_base_url}/3115-maximum-prime-difference.adoc[题解]
+|Medium
+|
 
 //|{counter:codes}
 //|{leetcode_base_url}/kth-smallest-amount-with-single-denomination-combination/[3116. Kth Smallest Amount With Single Denomination Combination^]

docs/3115-maximum-prime-difference.adoc

@@ -1,46 +1,42 @@
 [#3115-maximum-prime-difference]
-= 3115. Maximum Prime Difference
+= 3115. 质数的最大距离
 
-{leetcode}/problems/maximum-prime-difference/[LeetCode - 3115. Maximum Prime Difference ^]
+https://leetcode.cn/problems/maximum-prime-difference/[LeetCode - 3115. 质数的最大距离 ^]
 
-You are given an integer array `nums`.
+给你一个整数数组 `nums`。
 
-Return an integer that is the *maximum* distance between the *indices* of two (not necessarily different) prime numbers in `nums`_._
+返回两个（不一定不同的）质数在 `nums` 中 *下标* 的 *最大距离*。
 
- 
-*Example 1:*
+*示例 1：*
 
-<div class="example-block">
-*Input:* <span class="example-io">nums = [4,2,9,5,3]
+****
+*输入：* [.example-io]#nums = [4,2,9,5,3]#
 
-*Output:* <span class="example-io">3
+*输出：* [.example-io]#3#
 
-*Explanation:* `nums[1]`, `nums[3]`, and `nums[4]` are prime. So the answer is `|4 - 1| = 3`.
+*解释：* `nums[1]`、`nums[3]` 和 `nums[4]` 是质数。因此答案是 `|4 - 1| = 3`。
+****
 
+*示例 2：*
 
-*Example 2:*
+****
+*输入：* [.example-io]#nums = [4,8,2,8]#
 
-<div class="example-block">
-*Input:* <span class="example-io">nums = [4,8,2,8]
+*输出：* [.example-io]#0#
 
-*Output:* <span class="example-io">0
-
-*Explanation:* `nums[2]` is prime. Because there is just one prime number, the answer is `|2 - 2| = 0`.
-
-
- 
-*Constraints:*
-
-
-* `1 <= nums.length <= 3 * 10^5^`
-* `1 <= nums[i] <= 100`
-* The input is generated such that the number of prime numbers in the `nums` is at least one.
+*解释：* `nums[2]` 是质数。因为只有一个质数，所以答案是 `|2 - 2| = 0`。
+****
 
+*提示：*
 
+* `1 \<= nums.length \<= 3 * 10^5^`
+* `+1 <= nums[i] <= 100+`
+* 输入保证 `nums` 中至少有一个质数。
 
 
 == 思路分析
 
+从数组前后分别寻找质数，然后坐标相减即可。
 
 [[src-3115]]
 [tabs]

docs/index.adoc

@@ -6314,7 +6314,7 @@ include::3099-harshad-number.adoc[leveloffset=+1]
 
 // include::3114-latest-time-you-can-obtain-after-replacing-characters.adoc[leveloffset=+1]
 
-// include::3115-maximum-prime-difference.adoc[leveloffset=+1]
+include::3115-maximum-prime-difference.adoc[leveloffset=+1]
 
 // include::3116-kth-smallest-amount-with-single-denomination-combination.adoc[leveloffset=+1]
 

logbook/202503.adoc

@@ -638,6 +638,11 @@ endif::[]
 |{doc_base_url}/0144-binary-tree-preorder-traversal.adoc[题解]
 |✅ 树的 Morris 遍历
 
+|{counter:codes2503}
+|{leetcode_base_url}/maximum-prime-difference/[3115. 质数的最大距离^]
+|{doc_base_url}/3115-maximum-prime-difference.adoc[题解]
+|✅ 质数判断
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_3115_MaximumPrimeDifference.java

@@ -0,0 +1,46 @@
+package com.diguage.algo.leetcode;
+
+public class _3115_MaximumPrimeDifference {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-05-09 21:44:53
+   */
+  public int maximumPrimeDifference(int[] nums) {
+    int min = -1;
+    for (int i = 0; i < nums.length; i++) {
+      if (isPrime(nums[i])) {
+        min = i;
+        break;
+      }
+    }
+    int max = nums.length;
+    for (int i = nums.length - 1; i >= min; i--) {
+      if (isPrime(nums[i])) {
+        max = i;
+        break;
+      }
+    }
+    return max - min;
+  }
+
+  private boolean isPrime(int num) {
+    if (num < 2) {
+      return false;
+    }
+    if (num == 2) {
+      return true;
+    }
+    if ((num & 1) == 0) {
+      return false;
+    }
+    for (int i = 2; i * i <= num; i++) {
+      if (num % i == 0) {
+        return false;
+      }
+    }
+    return true;
+  }
+  // end::answer[]
+}