feat: add solutions to lc problem: No.2110 (#4905)

Author: yanglbme

solution/2100-2199/2110.Number of Smooth Descent Periods of a Stock/README.md

@@ -69,15 +69,15 @@ tags:
 
 ### 方法一：双指针
 
-我们定义一个答案变量 `ans`，初始值为 $0$。
+我们定义一个答案变量 $\textit{ans}$，初始值为 $0$。
 
 接下来，我们使用双指针 $i$ 和 $j$，分别指向当前平滑下降阶段的第一天和最后一天的下一天。初始时 $i = 0$, $j = 0$。
 
-从左到右遍历数组 `prices`，对于每个位置 $i$，我们将 $j$ 向右移动，直到 $j$ 到达数组末尾或者 $prices[j - 1] - prices[j] \neq 1$ 为止。此时，$cnt = j - i$ 即为当前平滑下降阶段的长度，我们累加 $\frac{(1 + cnt) \times cnt}{2}$ 到答案变量 `ans` 中。接下来将 $i$ 更新为 $j$，继续遍历。
+从左到右遍历数组 $\textit{prices}$，对于每个位置 $i$，我们将 $j$ 向右移动，直到 $j$ 到达数组末尾或者 $\textit{prices}[j - 1] - \textit{prices}[j] \neq 1$ 为止。此时，$\textit{cnt} = j - i$ 即为当前平滑下降阶段的长度，我们累加 $\frac{(1 + \textit{cnt}) \times \textit{cnt}}{2}$ 到答案变量 $\textit{ans}$ 中。接下来将 $i$ 更新为 $j$，继续遍历。
 
-遍历结束后，返回答案变量 `ans` 即可。
+遍历结束后，返回答案变量 $\textit{ans}$ 即可。
 
-时间复杂度 $O(n)$，其中 $n$ 为数组 `prices` 的长度。空间复杂度 $O(1)$。
+时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{prices}$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -174,6 +174,50 @@ function getDescentPeriods(prices: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn get_descent_periods(prices: Vec<i32>) -> i64 {
+        let mut ans: i64 = 0;
+        let n: usize = prices.len();
+        let mut i: usize = 0;
+
+        while i < n {
+            let mut j: usize = i + 1;
+            while j < n && prices[j - 1] - prices[j] == 1 {
+                j += 1;
+            }
+            let cnt: i64 = (j - i) as i64;
+            ans += (1 + cnt) * cnt / 2;
+            i = j;
+        }
+
+        ans
+    }
+}
+```
+
+#### C#
+
+```cs
+public class Solution {
+    public long GetDescentPeriods(int[] prices) {
+        long ans = 0;
+        int n = prices.Length;
+        for (int i = 0, j = 0; i < n; i = j) {
+            j = i + 1;
+            while (j < n && prices[j - 1] - prices[j] == 1) {
+                ++j;
+            }
+            int cnt = j - i;
+            ans += (1L + cnt) * cnt / 2;
+        }
+        return ans;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2100-2199/2110.Number of Smooth Descent Periods of a Stock/README_EN.md

@@ -70,15 +70,15 @@ Note that [8,6] is not a smooth descent period as 8 - 6 ≠ 1.
 
 ### Solution 1: Two Pointers
 
-We define an answer variable `ans`, initially set to $0$.
+We define an answer variable $\textit{ans}$ with an initial value of $0$.
 
-Next, we use two pointers $i$ and $j$, pointing to the first day of the current smooth decline phase and the day after the last day of this phase, respectively. Initially, $i = 0$, $j = 0$.
+Next, we use two pointers $i$ and $j$, which point to the first day of the current smooth descent period and the day after the last day, respectively. Initially, $i = 0$ and $j = 0$.
 
-Iterate through the array `prices` from left to right. For each position $i$, we move $j$ to the right until $j$ reaches the end of the array or $prices[j - 1] - prices[j] \neq 1$. At this point, $cnt = j - i$ is the length of the current smooth decline phase, and we add $\frac{(1 + cnt) \times cnt}{2}$ to the answer variable `ans`. Then, we update $i$ to $j$ and continue the iteration.
+We traverse the array $\textit{prices}$ from left to right. For each position $i$, we move $j$ to the right until $j$ reaches the end of the array or $\textit{prices}[j - 1] - \textit{prices}[j] \neq 1$. At this point, $\textit{cnt} = j - i$ is the length of the current smooth descent period, and we add $\frac{(1 + \textit{cnt}) \times \textit{cnt}}{2}$ to the answer variable $\textit{ans}$. Then we update $i$ to $j$ and continue traversing.
 
-After the iteration ends, return the answer variable `ans`.
+After the traversal ends, we return the answer variable $\textit{ans}$.
 
-The time complexity is $O(n)$, where $n$ is the length of the array `prices`. The space complexity is $O(1)$.
+The time complexity is $O(n)$, where $n$ is the length of the array $\textit{prices}$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -175,6 +175,50 @@ function getDescentPeriods(prices: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn get_descent_periods(prices: Vec<i32>) -> i64 {
+        let mut ans: i64 = 0;
+        let n: usize = prices.len();
+        let mut i: usize = 0;
+
+        while i < n {
+            let mut j: usize = i + 1;
+            while j < n && prices[j - 1] - prices[j] == 1 {
+                j += 1;
+            }
+            let cnt: i64 = (j - i) as i64;
+            ans += (1 + cnt) * cnt / 2;
+            i = j;
+        }
+
+        ans
+    }
+}
+```
+
+#### C#
+
+```cs
+public class Solution {
+    public long GetDescentPeriods(int[] prices) {
+        long ans = 0;
+        int n = prices.Length;
+        for (int i = 0, j = 0; i < n; i = j) {
+            j = i + 1;
+            while (j < n && prices[j - 1] - prices[j] == 1) {
+                ++j;
+            }
+            int cnt = j - i;
+            ans += (1L + cnt) * cnt / 2;
+        }
+        return ans;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2100-2199/2110.Number of Smooth Descent Periods of a Stock/Solution.cs

@@ -0,0 +1,15 @@
+public class Solution {
+    public long GetDescentPeriods(int[] prices) {
+        long ans = 0;
+        int n = prices.Length;
+        for (int i = 0, j = 0; i < n; i = j) {
+            j = i + 1;
+            while (j < n && prices[j - 1] - prices[j] == 1) {
+                ++j;
+            }
+            int cnt = j - i;
+            ans += (1L + cnt) * cnt / 2;
+        }
+        return ans;
+    }
+}

solution/2100-2199/2110.Number of Smooth Descent Periods of a Stock/Solution.rs

@@ -0,0 +1,19 @@
+impl Solution {
+    pub fn get_descent_periods(prices: Vec<i32>) -> i64 {
+        let mut ans: i64 = 0;
+        let n: usize = prices.len();
+        let mut i: usize = 0;
+
+        while i < n {
+            let mut j: usize = i + 1;
+            while j < n && prices[j - 1] - prices[j] == 1 {
+                j += 1;
+            }
+            let cnt: i64 = (j - i) as i64;
+            ans += (1 + cnt) * cnt / 2;
+            i = j;
+        }
+
+        ans
+    }
+}