From 5ebff9327ddbb9abf9617f2e973e4661ff3ad3f7 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Wed, 22 Jan 2025 12:55:14 +0800 Subject: [PATCH] feat: add solutions to lc problem: No.2419 No.2419.Longest Subarray With Maximum Bitwise AND --- .../README.md | 42 ++------- .../README_EN.md | 40 ++------- .../Solution.cpp | 6 +- .../Solution.js | 6 +- .../Solution.ts | 6 +- .../Solution2.ts | 3 - .../README.md | 82 +++++++++++------- .../README_EN.md | 86 +++++++++++-------- .../Solution.cpp | 11 ++- .../Solution.go | 12 +-- .../Solution.java | 14 ++- .../Solution.js | 13 +-- .../Solution.py | 4 +- .../Solution.rs | 18 ++++ .../Solution.ts | 5 +- 15 files changed, 167 insertions(+), 181 deletions(-) delete mode 100644 solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution2.ts create mode 100644 solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.rs diff --git a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md index fb25fd9622540..d3c8e340c3d81 100644 --- a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md +++ b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md @@ -50,7 +50,7 @@ X++：X 加 1 ，X = 0 + 1 = 1 
 输入：operations = ["++X","++X","X++"]
 输出：3
-解释：操作按下述步骤执行： 
+解释：操作按下述步骤执行：
 最初，X = 0
 ++X：X 加 1 ，X = 0 + 1 = 1
 ++X：X 加 1 ，X = 1 + 1 = 2
@@ -85,11 +85,11 @@ X--：X 减 1 ，X = 1 - 1 = 0
 
 
 
-### 方法一：模拟
+### 方法一：计数
 
-遍历数组 `operations`，对于每个操作 $operations[i]$，如果包含 `'+'`，那么答案加 $1$，否则答案减 $1$。
+我们遍历数组 $\textit{operations}$，对于每个操作 $\textit{operations}[i]$，如果包含 `'+'`，那么答案加 $1$，否则答案减 $1$。
 
-时间复杂度为 $O(n)$，其中 $n$ 为数组 `operations` 的长度。空间复杂度 $O(1)$。
+时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{operations}$ 的长度。空间复杂度 $O(1)$。
 
 
 
@@ -122,7 +122,9 @@ class Solution {
 public:
     int finalValueAfterOperations(vector& operations) {
         int ans = 0;
-        for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
+        for (auto& s : operations) {
+            ans += s[1] == '+' ? 1 : -1;
+        }
         return ans;
     }
 };
@@ -147,11 +149,7 @@ func finalValueAfterOperations(operations []string) (ans int) {
 
 ```ts
 function finalValueAfterOperations(operations: string[]): number {
-    let ans = 0;
-    for (let operation of operations) {
-        ans += operation.includes('+') ? 1 : -1;
-    }
-    return ans;
+    return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
 }
 ```
 
@@ -177,11 +175,7 @@ impl Solution {
  * @return {number}
  */
 var finalValueAfterOperations = function (operations) {
-    let ans = 0;
-    for (const s of operations) {
-        ans += s[1] === '+' ? 1 : -1;
-    }
-    return ans;
+    return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
 };
 ```
 
@@ -201,22 +195,4 @@ int finalValueAfterOperations(char** operations, int operationsSize) {
 
 
 
-
-
-### 方法二
-
-
-
-#### TypeScript
-
-```ts
-function finalValueAfterOperations(operations: string[]): number {
-    return operations.reduce((r, v) => r + (v[1] === '+' ? 1 : -1), 0);
-}
-```
-
-
-
-
-
 
diff --git a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README_EN.md b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README_EN.md
index ade0b3ec54916..6a28786b3e0a3 100644
--- a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README_EN.md	
+++ b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/README_EN.md	
@@ -83,11 +83,11 @@ X--: X is decremented by 1, X = 1 - 1 = 0.
 
 
 
-### Solution 1: Simulation
+### Solution 1: Counting
 
-Traverse the array `operations`. For each operation $operations[i]$, if it contains `'+'`, then the answer increases by $1$, otherwise the answer decreases by $1$.
+We traverse the array $\textit{operations}$. For each operation $\textit{operations}[i]$, if it contains `'+'`, we increment the answer by $1$, otherwise, we decrement the answer by $1$.
 
-The time complexity is $O(n)$, where $n$ is the length of the array `operations`. The space complexity is $O(1)$.
+The time complexity is $O(n)$, where $n$ is the length of the array $\textit{operations}$. The space complexity is $O(1)$.
 
 
 
@@ -120,7 +120,9 @@ class Solution {
 public:
     int finalValueAfterOperations(vector& operations) {
         int ans = 0;
-        for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
+        for (auto& s : operations) {
+            ans += s[1] == '+' ? 1 : -1;
+        }
         return ans;
     }
 };
@@ -145,11 +147,7 @@ func finalValueAfterOperations(operations []string) (ans int) {
 
 ```ts
 function finalValueAfterOperations(operations: string[]): number {
-    let ans = 0;
-    for (let operation of operations) {
-        ans += operation.includes('+') ? 1 : -1;
-    }
-    return ans;
+    return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
 }
 ```
 
@@ -175,11 +173,7 @@ impl Solution {
  * @return {number}
  */
 var finalValueAfterOperations = function (operations) {
-    let ans = 0;
-    for (const s of operations) {
-        ans += s[1] === '+' ? 1 : -1;
-    }
-    return ans;
+    return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
 };
 ```
 
@@ -199,22 +193,4 @@ int finalValueAfterOperations(char** operations, int operationsSize) {
 
 
 
-
-
-### Solution 2
-
-
-
-#### TypeScript
-
-```ts
-function finalValueAfterOperations(operations: string[]): number {
-    return operations.reduce((r, v) => r + (v[1] === '+' ? 1 : -1), 0);
-}
-```
-
-
-
-
-
 
diff --git a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.cpp b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.cpp
index 7e05fc3bd911a..75fb624cc8acc 100644
--- a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.cpp	
+++ b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.cpp	
@@ -2,7 +2,9 @@ class Solution {
 public:
     int finalValueAfterOperations(vector& operations) {
         int ans = 0;
-        for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
+        for (auto& s : operations) {
+            ans += s[1] == '+' ? 1 : -1;
+        }
         return ans;
     }
-};
\ No newline at end of file
+};
diff --git a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.js b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.js
index c23c95d361e84..02583d004cbe2 100644
--- a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.js	
+++ b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.js	
@@ -3,9 +3,5 @@
  * @return {number}
  */
 var finalValueAfterOperations = function (operations) {
-    let ans = 0;
-    for (const s of operations) {
-        ans += s[1] === '+' ? 1 : -1;
-    }
-    return ans;
+    return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
 };
diff --git a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.ts b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.ts
index acc0823407810..3595a617d6b1e 100644
--- a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.ts	
+++ b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.ts	
@@ -1,7 +1,3 @@
 function finalValueAfterOperations(operations: string[]): number {
-    let ans = 0;
-    for (let operation of operations) {
-        ans += operation.includes('+') ? 1 : -1;
-    }
-    return ans;
+    return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
 }
diff --git a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution2.ts b/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution2.ts
deleted file mode 100644
index c5c55d14ccd3e..0000000000000
--- a/solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution2.ts	
+++ /dev/null
@@ -1,3 +0,0 @@
-function finalValueAfterOperations(operations: string[]): number {
-    return operations.reduce((r, v) => r + (v[1] === '+' ? 1 : -1), 0);
-}
diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README.md b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README.md
index db36e13eca450..38ace23072d25 100644
--- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README.md	
+++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README.md	
@@ -52,7 +52,7 @@ tags:
 输入：nums = [1,2,3,4]
 输出：1
 解释：
-子数组按位与运算的最大值是 4 。 
+子数组按位与运算的最大值是 4 。
 能得到此结果的最长子数组是 [4]，所以返回 1 。
@@ -77,9 +77,9 @@ tags: 题目可以转换为求最大值在数组中最多连续出现的次数。 -先遍历一遍数组，求出最大值，然后再遍历一遍数组，求出最大值连续出现的次数，最后返回这个次数即可。 +我们先遍历数组 $\textit{nums}$ 找到最大值 $\textit{mx}$，然后再遍历数组一次，找到最大值连续出现的次数，最后返回这个次数即可。 -时间复杂度 $O(n)$。其中 $n$ 为数组的长度。 +时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。 @@ -90,8 +90,8 @@ class Solution: def longestSubarray(self, nums: List[int]) -> int: mx = max(nums) ans = cnt = 0 - for v in nums: - if v == mx: + for x in nums: + if x == mx: cnt += 1 ans = max(ans, cnt) else: @@ -104,15 +104,11 @@ class Solution: ```java class Solution { public int longestSubarray(int[] nums) { - int mx = 0; - for (int v : nums) { - mx = Math.max(mx, v); - } + int mx = Arrays.stream(nums).max().getAsInt(); int ans = 0, cnt = 0; - for (int v : nums) { - if (v == mx) { - ++cnt; - ans = Math.max(ans, cnt); + for (int x : nums) { + if (x == mx) { + ans = Math.max(ans, ++cnt); } else { cnt = 0; } @@ -128,12 +124,11 @@ class Solution { class Solution { public: int longestSubarray(vector& nums) { - int mx = *max_element(nums.begin(), nums.end()); + int mx = ranges::max(nums); int ans = 0, cnt = 0; - for (int v : nums) { - if (v == mx) { - ++cnt; - ans = max(ans, cnt); + for (int x : nums) { + if (x == mx) { + ans = max(ans, ++cnt); } else { cnt = 0; } @@ -146,18 +141,18 @@ public: #### Go ```go -func longestSubarray(nums []int) int { +func longestSubarray(nums []int) (ans int) { mx := slices.Max(nums) - ans, cnt := 0, 0 - for _, v := range nums { - if v == mx { + cnt := 0 + for _, x := range nums { + if x == mx { cnt++ ans = max(ans, cnt) } else { cnt = 0 } } - return ans + return } ``` @@ -167,38 +162,59 @@ func longestSubarray(nums []int) int { function longestSubarray(nums: number[]): number { const mx = Math.max(...nums); let [ans, cnt] = [0, 0]; - for (const x of nums) { if (x === mx) { - cnt++; - ans = Math.max(ans, cnt); + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } - return ans; } ``` +#### Rust + +```rust +impl Solution { + pub fn longest_subarray(nums: Vec) -> i32 { + let mx = *nums.iter().max().unwrap(); + let mut ans = 0; + let mut cnt = 0; + + for &x in nums.iter() { + if x == mx { + cnt += 1; + ans = ans.max(cnt); + } else { + cnt = 0; + } + } + + ans + } +} +``` + #### JavaScript ```js -function longestSubarray(nums) { +/** + * @param {number[]} nums + * @return {number} + */ +var longestSubarray = function (nums) { const mx = Math.max(...nums); let [ans, cnt] = [0, 0]; - for (const x of nums) { if (x === mx) { - cnt++; - ans = Math.max(ans, cnt); + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } - return ans; -} +}; ``` diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README_EN.md b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README_EN.md index 83dd92a8226b9..c95729b44a910 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README_EN.md +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README_EN.md @@ -69,15 +69,15 @@ The longest subarray with that value is [4], so we return 1. -### Solution 1: Quick Thinking +### Solution 1: Brain Teaser -Due to the bitwise AND operation, the number will not get larger, so the maximum value is the maximum value in the array. +Since the bitwise AND operation does not increase the number, the maximum value is the maximum value in the array. -The problem can be transformed into finding the maximum number of consecutive occurrences of the maximum value in the array. +The problem can be converted to finding the maximum number of consecutive occurrences of the maximum value in the array. -First, traverse the array once to find the maximum value, then traverse the array again to find the number of consecutive occurrences of the maximum value, and finally return this count. +First, traverse the array $\textit{nums}$ to find the maximum value $\textit{mx}$, then traverse the array again to find the maximum number of consecutive occurrences of the maximum value. Finally, return this count. -The time complexity is $O(n)$, where $n$ is the length of the array. +The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$. @@ -88,8 +88,8 @@ class Solution: def longestSubarray(self, nums: List[int]) -> int: mx = max(nums) ans = cnt = 0 - for v in nums: - if v == mx: + for x in nums: + if x == mx: cnt += 1 ans = max(ans, cnt) else: @@ -102,15 +102,11 @@ class Solution: ```java class Solution { public int longestSubarray(int[] nums) { - int mx = 0; - for (int v : nums) { - mx = Math.max(mx, v); - } + int mx = Arrays.stream(nums).max().getAsInt(); int ans = 0, cnt = 0; - for (int v : nums) { - if (v == mx) { - ++cnt; - ans = Math.max(ans, cnt); + for (int x : nums) { + if (x == mx) { + ans = Math.max(ans, ++cnt); } else { cnt = 0; } @@ -126,12 +122,11 @@ class Solution { class Solution { public: int longestSubarray(vector& nums) { - int mx = *max_element(nums.begin(), nums.end()); + int mx = ranges::max(nums); int ans = 0, cnt = 0; - for (int v : nums) { - if (v == mx) { - ++cnt; - ans = max(ans, cnt); + for (int x : nums) { + if (x == mx) { + ans = max(ans, ++cnt); } else { cnt = 0; } @@ -144,18 +139,18 @@ public: #### Go ```go -func longestSubarray(nums []int) int { +func longestSubarray(nums []int) (ans int) { mx := slices.Max(nums) - ans, cnt := 0, 0 - for _, v := range nums { - if v == mx { + cnt := 0 + for _, x := range nums { + if x == mx { cnt++ ans = max(ans, cnt) } else { cnt = 0 } } - return ans + return } ``` @@ -165,38 +160,59 @@ func longestSubarray(nums []int) int { function longestSubarray(nums: number[]): number { const mx = Math.max(...nums); let [ans, cnt] = [0, 0]; - for (const x of nums) { if (x === mx) { - cnt++; - ans = Math.max(ans, cnt); + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } - return ans; } ``` +#### Rust + +```rust +impl Solution { + pub fn longest_subarray(nums: Vec) -> i32 { + let mx = *nums.iter().max().unwrap(); + let mut ans = 0; + let mut cnt = 0; + + for &x in nums.iter() { + if x == mx { + cnt += 1; + ans = ans.max(cnt); + } else { + cnt = 0; + } + } + + ans + } +} +``` + #### JavaScript ```js -function longestSubarray(nums) { +/** + * @param {number[]} nums + * @return {number} + */ +var longestSubarray = function (nums) { const mx = Math.max(...nums); let [ans, cnt] = [0, 0]; - for (const x of nums) { if (x === mx) { - cnt++; - ans = Math.max(ans, cnt); + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } - return ans; -} +}; ``` diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.cpp b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.cpp index 257bd360db5d9..4150badd4cfb5 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.cpp +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.cpp @@ -1,16 +1,15 @@ class Solution { public: int longestSubarray(vector& nums) { - int mx = *max_element(nums.begin(), nums.end()); + int mx = ranges::max(nums); int ans = 0, cnt = 0; - for (int v : nums) { - if (v == mx) { - ++cnt; - ans = max(ans, cnt); + for (int x : nums) { + if (x == mx) { + ans = max(ans, ++cnt); } else { cnt = 0; } } return ans; } -}; \ No newline at end of file +}; diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.go b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.go index b0a6a2dc19d18..90851d9be9877 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.go +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.go @@ -1,13 +1,13 @@ -func longestSubarray(nums []int) int { +func longestSubarray(nums []int) (ans int) { mx := slices.Max(nums) - ans, cnt := 0, 0 - for _, v := range nums { - if v == mx { + cnt := 0 + for _, x := range nums { + if x == mx { cnt++ ans = max(ans, cnt) } else { cnt = 0 } } - return ans -} \ No newline at end of file + return +} diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.java b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.java index 00970ca639bd8..655523166ca00 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.java +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.java @@ -1,18 +1,14 @@ class Solution { public int longestSubarray(int[] nums) { - int mx = 0; - for (int v : nums) { - mx = Math.max(mx, v); - } + int mx = Arrays.stream(nums).max().getAsInt(); int ans = 0, cnt = 0; - for (int v : nums) { - if (v == mx) { - ++cnt; - ans = Math.max(ans, cnt); + for (int x : nums) { + if (x == mx) { + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } return ans; } -} \ No newline at end of file +} diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.js b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.js index c949ce8d97802..05978ce953fdb 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.js +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.js @@ -1,15 +1,16 @@ -function longestSubarray(nums) { +/** + * @param {number[]} nums + * @return {number} + */ +var longestSubarray = function (nums) { const mx = Math.max(...nums); let [ans, cnt] = [0, 0]; - for (const x of nums) { if (x === mx) { - cnt++; - ans = Math.max(ans, cnt); + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } - return ans; -} +}; diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.py b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.py index 490d34af6eeac..3e916dad58a08 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.py +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.py @@ -2,8 +2,8 @@ class Solution: def longestSubarray(self, nums: List[int]) -> int: mx = max(nums) ans = cnt = 0 - for v in nums: - if v == mx: + for x in nums: + if x == mx: cnt += 1 ans = max(ans, cnt) else: diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.rs b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.rs new file mode 100644 index 0000000000000..62e5eb18f3749 --- /dev/null +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.rs @@ -0,0 +1,18 @@ +impl Solution { + pub fn longest_subarray(nums: Vec) -> i32 { + let mx = *nums.iter().max().unwrap(); + let mut ans = 0; + let mut cnt = 0; + + for &x in nums.iter() { + if x == mx { + cnt += 1; + ans = ans.max(cnt); + } else { + cnt = 0; + } + } + + ans + } +} diff --git a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.ts b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.ts index 4bd008ce7efba..149ff6a1b27a8 100644 --- a/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.ts +++ b/solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/Solution.ts @@ -1,15 +1,12 @@ function longestSubarray(nums: number[]): number { const mx = Math.max(...nums); let [ans, cnt] = [0, 0]; - for (const x of nums) { if (x === mx) { - cnt++; - ans = Math.max(ans, cnt); + ans = Math.max(ans, ++cnt); } else { cnt = 0; } } - return ans; }