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Commit fbd11e4

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feat: update solutions (doocs#2258)
1 parent 47f059e commit fbd11e4

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6 files changed

+28
-46
lines changed

6 files changed

+28
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‎solution/0000-0099/0088.Merge Sorted Array/README.md‎

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@@ -131,6 +131,18 @@ function merge(nums1: number[], m: number, nums2: number[], n: number): void {
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}
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```
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```ts
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/**
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Do not return anything, modify nums1 in-place instead.
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*/
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function merge(nums1: number[], m: number, nums2: number[], n: number): void {
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nums1.length = m;
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nums2.length = n;
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nums1.push(...nums2);
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nums1.sort((a, b) => a - b);
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}
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```
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```rust
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impl Solution {
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pub fn merge(nums1: &mut Vec<i32>, m: i32, nums2: &mut Vec<i32>, n: i32) {
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<!-- tabs:end -->
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### 方法二
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<!-- tabs:start -->
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```ts
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/**
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Do not return anything, modify nums1 in-place instead.
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*/
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function merge(nums1: number[], m: number, nums2: number[], n: number): void {
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nums1.length = m;
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nums2.length = n;
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nums1.push(...nums2);
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nums1.sort((a, b) => a - b);
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}
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```
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<!-- tabs:end -->
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<!-- end -->

‎solution/0000-0099/0088.Merge Sorted Array/README_EN.md‎

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@@ -126,6 +126,18 @@ function merge(nums1: number[], m: number, nums2: number[], n: number): void {
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}
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```
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```ts
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/**
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Do not return anything, modify nums1 in-place instead.
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*/
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function merge(nums1: number[], m: number, nums2: number[], n: number): void {
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nums1.length = m;
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nums2.length = n;
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nums1.push(...nums2);
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nums1.sort((a, b) => a - b);
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}
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```
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```rust
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impl Solution {
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pub fn merge(nums1: &mut Vec<i32>, m: i32, nums2: &mut Vec<i32>, n: i32) {
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<!-- tabs:end -->
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### Solution 2
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<!-- tabs:start -->
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```ts
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/**
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Do not return anything, modify nums1 in-place instead.
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*/
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function merge(nums1: number[], m: number, nums2: number[], n: number): void {
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nums1.length = m;
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nums2.length = n;
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nums1.push(...nums2);
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nums1.sort((a, b) => a - b);
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}
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```
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<!-- tabs:end -->
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<!-- end -->

‎solution/0000-0099/0091.Decode Ways/README.md‎

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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是字符串的长度。
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我们注意到,状态 $f[i]$ 仅与状态 $f[i-1]$ 和状态 $f[i-2]$ 有关,而与其他状态无关,因此我们可以使用两个变量代替这两个状态,使得原来的空间复杂度 $O(n)$ 降低至 $O(1)$。
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<!-- tabs:start -->
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```python
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<!-- tabs:end -->
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### 方法二
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我们注意到,状态 $f[i]$ 仅与状态 $f[i-1]$ 和状态 $f[i-2]$ 有关,而与其他状态无关,因此我们可以使用两个变量代替这两个状态,使得原来的空间复杂度 $O(n)$ 降低至 $O(1)$。
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<!-- tabs:start -->
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‎solution/0000-0099/0091.Decode Ways/README_EN.md‎

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<!-- tabs:end -->
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### Solution 2
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We notice that the state $f[i]$ is only related to the states $f[i-1]$ and $f[i-2],ドル and is irrelevant to other states. Therefore, we can use two variables to replace these two states, reducing the original space complexity from $O(n)$ to $O(1)$.
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<!-- tabs:start -->
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‎solution/0000-0099/0097.Interleaving String/README.md‎

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时间复杂度 $O(m \times n),ドル空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是字符串 $s_1$ 和 $s_2$ 的长度。
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我们注意到,状态 $f[i][j]$ 只和状态 $f[i - 1][j]$、$f[i][j - 1]$、$f[i - 1][j - 1]$ 有关,因此我们可以使用滚动数组优化空间复杂度,将空间复杂度优化到 $O(n)$。
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<!-- tabs:start -->
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```python
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<!-- tabs:end -->
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### 方法三
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我们注意到,状态 $f[i][j]$ 只和状态 $f[i - 1][j]$、$f[i][j - 1]$、$f[i - 1][j - 1]$ 有关,因此我们可以使用滚动数组优化空间复杂度,将空间复杂度优化到 $O(n)$。
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<!-- tabs:start -->
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‎solution/0000-0099/0097.Interleaving String/README_EN.md‎

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@@ -357,8 +357,6 @@ The answer is $f[m][n]$.
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The time complexity is $O(m \times n),ドル and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of strings $s_1$ and $s_2$ respectively.
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We notice that the state $f[i][j]$ is only related to the states $f[i - 1][j],ドル $f[i][j - 1],ドル and $f[i - 1][j - 1]$. Therefore, we can use a rolling array to optimize the space complexity, reducing the original space complexity from $O(m \times n)$ to $O(n)$.
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<!-- tabs:start -->
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```python
@@ -508,7 +506,7 @@ public class Solution {
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<!-- tabs:end -->
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### Solution 3
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We notice that the state $f[i][j]$ is only related to the states $f[i - 1][j],ドル $f[i][j - 1],ドル and $f[i - 1][j - 1]$. Therefore, we can use a rolling array to optimize the space complexity, reducing the original space complexity from $O(m \times n)$ to $O(n)$.
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<!-- tabs:start -->
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