From 38817f57aa638f0679c341ce8b6844349912a590 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Tue, 4 Feb 2025 08:19:13 +0800 Subject: [PATCH] feat: add solutions to lc problem: No.0922 No.0922.Sort Array By Parity II --- .../0922.Sort Array By Parity II/README.md | 26 ++++++++++++++++--- .../0922.Sort Array By Parity II/README_EN.md | 26 ++++++++++++++++--- .../0922.Sort Array By Parity II/Solution.rs | 15 +++++++++++ 3 files changed, 61 insertions(+), 6 deletions(-) create mode 100644 solution/0900-0999/0922.Sort Array By Parity II/Solution.rs diff --git a/solution/0900-0999/0922.Sort Array By Parity II/README.md b/solution/0900-0999/0922.Sort Array By Parity II/README.md index 02336d6b94a1d..4ab895ae65250 100644 --- a/solution/0900-0999/0922.Sort Array By Parity II/README.md +++ b/solution/0900-0999/0922.Sort Array By Parity II/README.md @@ -64,11 +64,11 @@ tags: ### 方法一：双指针 -我们用两个指针 $i$ 和 $j$ 分别指向偶数下标和奇数下标。 +我们用两个指针 $i$ 和 $j$ 分别指向偶数下标和奇数下标，初始时 $i = 0$, $j = 1$。 -当 $i$ 指向偶数下标时，如果 $nums[i]$ 是奇数，那么我们需要找到一个奇数下标 $j$，使得 $nums[j]$ 是偶数，然后交换 $nums[i]$ 和 $nums[j]$。继续遍历，直到 $i$ 指向数组末尾。 +当 $i$ 指向偶数下标时，如果 $\textit{nums}[i]$ 是奇数，那么我们需要找到一个奇数下标 $j$，使得 $\textit{nums}[j]$ 是偶数，然后交换 $\textit{nums}[i]$ 和 $\textit{nums}[j]$。继续遍历，直到 $i$ 指向数组末尾。 -时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。 +时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}[i]$ 的长度。空间复杂度 $O(1)$。 @@ -157,6 +157,26 @@ function sortArrayByParityII(nums: number[]): number[] { } ``` +#### Rust + +```rust +impl Solution { + pub fn sort_array_by_parity_ii(mut nums: Vec) -> Vec { + let n = nums.len(); + let mut j = 1; + for i in (0..n).step_by(2) { + if nums[i] % 2 != 0 { + while nums[j] % 2 != 0 { + j += 2; + } + nums.swap(i, j); + } + } + nums + } +} +``` + #### JavaScript ```js diff --git a/solution/0900-0999/0922.Sort Array By Parity II/README_EN.md b/solution/0900-0999/0922.Sort Array By Parity II/README_EN.md index 6e26f6ac04a9d..5d4ab1f7840e4 100644 --- a/solution/0900-0999/0922.Sort Array By Parity II/README_EN.md +++ b/solution/0900-0999/0922.Sort Array By Parity II/README_EN.md @@ -61,11 +61,11 @@ tags: ### Solution 1: Two Pointers -We use two pointers $i$ and $j$ to point to even and odd indices respectively. +We use two pointers $i$ and $j$ to point to even and odd indices, respectively. Initially, $i = 0$ and $j = 1$. -When $i$ points to an even index, if $nums[i]$ is odd, then we need to find an odd index $j$ such that $nums[j]$ is even, and then swap $nums[i]$ and $nums[j]$. Continue to iterate until $i$ points to the end of the array. +When $i$ points to an even index, if $\textit{nums}[i]$ is odd, we need to find an odd index $j$ such that $\textit{nums}[j]$ is even, and then swap $\textit{nums}[i]$ and $\textit{nums}[j]$. Continue traversing until $i$ reaches the end of the array. -The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$. +The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$. @@ -154,6 +154,26 @@ function sortArrayByParityII(nums: number[]): number[] { } ``` +#### Rust + +```rust +impl Solution { + pub fn sort_array_by_parity_ii(mut nums: Vec) -> Vec { + let n = nums.len(); + let mut j = 1; + for i in (0..n).step_by(2) { + if nums[i] % 2 != 0 { + while nums[j] % 2 != 0 { + j += 2; + } + nums.swap(i, j); + } + } + nums + } +} +``` + #### JavaScript ```js diff --git a/solution/0900-0999/0922.Sort Array By Parity II/Solution.rs b/solution/0900-0999/0922.Sort Array By Parity II/Solution.rs new file mode 100644 index 0000000000000..07be9bae68ccf --- /dev/null +++ b/solution/0900-0999/0922.Sort Array By Parity II/Solution.rs @@ -0,0 +1,15 @@ +impl Solution { + pub fn sort_array_by_parity_ii(mut nums: Vec) -> Vec { + let n = nums.len(); + let mut j = 1; + for i in (0..n).step_by(2) { + if nums[i] % 2 != 0 { + while nums[j] % 2 != 0 { + j += 2; + } + nums.swap(i, j); + } + } + nums + } +}