From 9f61549b005a9ee4c4978ec2bcf650ce755c81ab Mon Sep 17 00:00:00 2001 From: Libin YANG Date: Sat, 22 Mar 2025 11:51:52 +0800 Subject: [PATCH] feat: add solutions to lc problem: No.0325 No.0325.Maximum Size Subarray Sum Equals k --- .../README.md | 84 +++++++++++++++++-- .../README_EN.md | 84 ++++++++++++++++++- .../Solution.cs | 18 ++++ .../Solution.js | 21 +++++ .../Solution.rs | 20 +++++ 5 files changed, 221 insertions(+), 6 deletions(-) create mode 100644 solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.cs create mode 100644 solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.js create mode 100644 solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.rs diff --git a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README.md b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README.md index 128bebe51d7b5..ab2116885424a 100644 --- a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README.md +++ b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README.md @@ -26,7 +26,7 @@ tags: 
 输入: nums = [1,-1,5,-2,3], k = 3
-输出: 4 
+输出: 4
 解释: 子数组 [1, -1, 5, -2] 和等于 3，且长度最长。
@@ -55,13 +55,13 @@ tags: ### 方法一：哈希表 + 前缀和 -我们可以用一个哈希表 $d$ 记录数组 $nums$ 中每个前缀和第一次出现的下标，初始时 $d[0] = -1$。另外定义一个变量 $s$ 记录前缀和。 +我们可以用一个哈希表 $\textit{d}$ 记录数组 $\textit{nums}$ 中每个前缀和第一次出现的下标，初始时 $\textit{d}[0] = -1$。另外定义一个变量 $\textit{s}$ 记录前缀和。 -接下来，遍历数组 $nums$，对于当前遍历到的数字 $nums[i]$，我们更新前缀和 $s = s + nums[i]$，如果 $s-k$ 在哈希表 $d$ 中存在，不妨记 $j = d[s - k]$，那么以 $nums[i]$ 结尾的符合条件的子数组的长度为 $i - j$，我们使用一个变量 $ans$ 来维护最长的符合条件的子数组的长度。然后，如果 $s$ 在哈希表中不存在，我们记录 $s$ 和对应的下标 $i$，即 $d[s] = i$，否则我们不更新 $d[s]$。需要注意的是，可能会有多个位置 $i$ 都满足 $s$ 的值，因此我们只记录最小的 $i$，这样就能保证子数组的长度最长。 +接下来，遍历数组 $\textit{nums}$，对于当前遍历到的数字 $\textit{nums}[i]$，我们更新前缀和 $\textit{s} = \textit{s} + \textit{nums}[i]$，如果 $\textit{s}-k$ 在哈希表 $\textit{d}$ 中存在，不妨记 $j = \textit{d}[\textit{s} - k]$，那么以 $\textit{nums}[i]$ 结尾的符合条件的子数组的长度为 $i - j$，我们使用一个变量 $\textit{ans}$ 来维护最长的符合条件的子数组的长度。然后，如果 $\textit{s}$ 在哈希表中不存在，我们记录 $\textit{s}$ 和对应的下标 $i$，即 $\textit{d}[\textit{s}] = i$，否则我们不更新 $\textit{d}[\textit{s}]$。需要注意的是，可能会有多个位置 $i$ 都满足 $\textit{s}$ 的值，因此我们只记录最小的 $i$，这样就能保证子数组的长度最长。 -遍历结束之后，我们返回 $ans$ 即可。 +遍历结束之后，我们返回 $\textit{ans}$ 即可。 -时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。 +时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。 @@ -163,6 +163,80 @@ function maxSubArrayLen(nums: number[], k: number): number { } ``` +#### Rust + +```rust +use std::collections::HashMap; + +impl Solution { + pub fn max_sub_array_len(nums: Vec, k: i32) -> i32 { + let mut d = HashMap::new(); + d.insert(0, -1); + let mut ans = 0; + let mut s = 0; + + for (i, &x) in nums.iter().enumerate() { + s += x; + if let Some(&j) = d.get(&(s - k)) { + ans = ans.max((i as i32) - j); + } + d.entry(s).or_insert(i as i32); + } + + ans + } +} +``` + +#### JavaScript + +```js +/** + * @param {number[]} nums + * @param {number} k + * @return {number} + */ +var maxSubArrayLen = function (nums, k) { + const d = new Map(); + d.set(0, -1); + let ans = 0; + let s = 0; + for (let i = 0; i < nums.length; ++i) { + s += nums[i]; + if (d.has(s - k)) { + ans = Math.max(ans, i - d.get(s - k)); + } + if (!d.has(s)) { + d.set(s, i); + } + } + return ans; +}; +``` + +#### C# + +```cs +public class Solution { + public int MaxSubArrayLen(int[] nums, int k) { + var d = new Dictionary(); + d[0] = -1; + int ans = 0; + int s = 0; + for (int i = 0; i < nums.Length; i++) { + s += nums[i]; + if (d.ContainsKey(s - k)) { + ans = Math.Max(ans, i - d[s - k]); + } + if (!d.ContainsKey(s)) { + d[s] = i; + } + } + return ans; + } +} +``` + diff --git a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README_EN.md b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README_EN.md index faa73742e4b73..6c708b0a1b002 100644 --- a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README_EN.md +++ b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/README_EN.md @@ -52,7 +52,15 @@ tags: -### Solution 1 +### Solution 1: Hash Table + Prefix Sum + +We can use a hash table $\textit{d}$ to record the first occurrence index of each prefix sum in the array $\textit{nums}$, initializing $\textit{d}[0] = -1$. Additionally, we define a variable $\textit{s}$ to keep track of the current prefix sum. + +Next, we iterate through the array $\textit{nums}$. For the current number $\textit{nums}[i]$, we update the prefix sum $\textit{s} = \textit{s} + \textit{nums}[i]$. If $\textit{s} - k$ exists in the hash table $\textit{d}$, let $\textit{j} = \textit{d}[\textit{s} - k]$, then the length of the subarray that ends at $\textit{nums}[i]$ and satisfies the condition is $i - j$. We use a variable $\textit{ans}$ to maintain the length of the longest subarray that satisfies the condition. After that, if $\textit{s}$ does not exist in the hash table, we record $\textit{s}$ and its corresponding index $i$ by setting $\textit{d}[\textit{s}] = i$. Otherwise, we do not update $\textit{d}[\textit{s}]$. It is important to note that there may be multiple positions $i$ with the same value of $\textit{s}$, so we only record the smallest $i$ to ensure the subarray length is the longest. + +After the iteration ends, we return $\textit{ans}$. + +The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. @@ -154,6 +162,80 @@ function maxSubArrayLen(nums: number[], k: number): number { } ``` +#### Rust + +```rust +use std::collections::HashMap; + +impl Solution { + pub fn max_sub_array_len(nums: Vec, k: i32) -> i32 { + let mut d = HashMap::new(); + d.insert(0, -1); + let mut ans = 0; + let mut s = 0; + + for (i, &x) in nums.iter().enumerate() { + s += x; + if let Some(&j) = d.get(&(s - k)) { + ans = ans.max((i as i32) - j); + } + d.entry(s).or_insert(i as i32); + } + + ans + } +} +``` + +#### JavaScript + +```js +/** + * @param {number[]} nums + * @param {number} k + * @return {number} + */ +var maxSubArrayLen = function (nums, k) { + const d = new Map(); + d.set(0, -1); + let ans = 0; + let s = 0; + for (let i = 0; i < nums.length; ++i) { + s += nums[i]; + if (d.has(s - k)) { + ans = Math.max(ans, i - d.get(s - k)); + } + if (!d.has(s)) { + d.set(s, i); + } + } + return ans; +}; +``` + +#### C# + +```cs +public class Solution { + public int MaxSubArrayLen(int[] nums, int k) { + var d = new Dictionary(); + d[0] = -1; + int ans = 0; + int s = 0; + for (int i = 0; i < nums.Length; i++) { + s += nums[i]; + if (d.ContainsKey(s - k)) { + ans = Math.Max(ans, i - d[s - k]); + } + if (!d.ContainsKey(s)) { + d[s] = i; + } + } + return ans; + } +} +``` + diff --git a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.cs b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.cs new file mode 100644 index 0000000000000..209eabe2162d3 --- /dev/null +++ b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.cs @@ -0,0 +1,18 @@ +public class Solution { + public int MaxSubArrayLen(int[] nums, int k) { + var d = new Dictionary(); + d[0] = -1; + int ans = 0; + int s = 0; + for (int i = 0; i < nums.Length; i++) { + s += nums[i]; + if (d.ContainsKey(s - k)) { + ans = Math.Max(ans, i - d[s - k]); + } + if (!d.ContainsKey(s)) { + d[s] = i; + } + } + return ans; + } +} diff --git a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.js b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.js new file mode 100644 index 0000000000000..dc542c5d2fa24 --- /dev/null +++ b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.js @@ -0,0 +1,21 @@ +/** + * @param {number[]} nums + * @param {number} k + * @return {number} + */ +var maxSubArrayLen = function (nums, k) { + const d = new Map(); + d.set(0, -1); + let ans = 0; + let s = 0; + for (let i = 0; i < nums.length; ++i) { + s += nums[i]; + if (d.has(s - k)) { + ans = Math.max(ans, i - d.get(s - k)); + } + if (!d.has(s)) { + d.set(s, i); + } + } + return ans; +}; diff --git a/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.rs b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.rs new file mode 100644 index 0000000000000..e011cb1e1c458 --- /dev/null +++ b/solution/0300-0399/0325.Maximum Size Subarray Sum Equals k/Solution.rs @@ -0,0 +1,20 @@ +use std::collections::HashMap; + +impl Solution { + pub fn max_sub_array_len(nums: Vec, k: i32) -> i32 { + let mut d = HashMap::new(); + d.insert(0, -1); + let mut ans = 0; + let mut s = 0; + + for (i, &x) in nums.iter().enumerate() { + s += x; + if let Some(&j) = d.get(&(s - k)) { + ans = ans.max((i as i32) - j); + } + d.entry(s).or_insert(i as i32); + } + + ans + } +}