From 6253f3c883496d81d0a87d92cf21da8b894701e4 Mon Sep 17 00:00:00 2001
From: Libin YANG <contact@yanglibin.info>
Date: Fri, 10 Oct 2025 07:01:03 +0800
Subject: [PATCH] feat: add solutions to lc problem: No.3147

---
 .../README.md                                 | 25 +++++++++++++++--
 .../README_EN.md                              | 27 ++++++++++++++++---
 .../Solution.rs                               | 16 +++++++++++
 3 files changed, 63 insertions(+), 5 deletions(-)
 create mode 100644 solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/Solution.rs

diff --git a/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README.md b/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README.md
index 5c21898309773..01b4a3fa27625 100644
--- a/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README.md	
+++ b/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README.md	
@@ -97,11 +97,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一：枚举 + 后缀和
+### 方法一：枚举 + 逆序遍历
 
 我们可以在 $[n - k, n)$ 的范围内枚举终点，然后从终点开始向前遍历，每次累加间隔为 $k$ 的魔法师的能量值，更新答案。
 
-时间复杂度 $O(n)$，其中 $n$ 是数组 `energy` 的长度。空间复杂度 $O(1)$。
+时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{energy}$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -190,6 +190,27 @@ function maximumEnergy(energy: number[], k: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_energy(energy: Vec<i32>, k: i32) -> i32 {
+        let n = energy.len();
+        let mut ans = i32::MIN;
+        for i in n - k as usize..n {
+            let mut s = 0;
+            let mut j = i as i32;
+            while j >= 0 {
+                s += energy[j as usize];
+                ans = ans.max(s);
+                j -= k;
+            }
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README_EN.md b/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README_EN.md
index d6a4d80e48d1c..b5e401697dcb4 100644
--- a/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README_EN.md	
+++ b/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README_EN.md	
@@ -98,11 +98,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1: Enumeration + Suffix Sum
+### Solution 1: Enumeration + Reverse Traversal
 
-We can enumerate the endpoint in the range of $[n - k, n)$, then start from the endpoint and traverse backwards, adding the energy value of the magician at each $k$ interval, and update the answer.
+We can enumerate the endpoints within the range $[n - k, n)$, then traverse backwards from each endpoint, accumulating the energy values of wizards at intervals of $k$, and update the answer.
 
-The time complexity is $O(n)$, where $n$ is the length of the array `energy`. The space complexity is $O(1)$.
+The time complexity is $O(n)$, where $n$ is the length of array $\textit{energy}$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -191,6 +191,27 @@ function maximumEnergy(energy: number[], k: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_energy(energy: Vec<i32>, k: i32) -> i32 {
+        let n = energy.len();
+        let mut ans = i32::MIN;
+        for i in n - k as usize..n {
+            let mut s = 0;
+            let mut j = i as i32;
+            while j >= 0 {
+                s += energy[j as usize];
+                ans = ans.max(s);
+                j -= k;
+            }
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/Solution.rs b/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/Solution.rs
new file mode 100644
index 0000000000000..cf501fd118328
--- /dev/null
+++ b/solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/Solution.rs	
@@ -0,0 +1,16 @@
+impl Solution {
+    pub fn maximum_energy(energy: Vec<i32>, k: i32) -> i32 {
+        let n = energy.len();
+        let mut ans = i32::MIN;
+        for i in n - k as usize..n {
+            let mut s = 0;
+            let mut j = i as i32;
+            while j >= 0 {
+                s += energy[j as usize];
+                ans = ans.max(s);
+                j -= k;
+            }
+        }
+        ans
+    }
+}