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feat: add solutions to lc problem: No.1441 #4268

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Mar 20, 2025
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feat: add solutions to lc problem: No.1441 #4268

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115 changes: 61 additions & 54 deletions solution/1400-1499/1441.Build an Array With Stack Operations/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -41,7 +41,7 @@ tags:
<pre>
<strong>输入：</strong>target = [1,3], n = 3
<strong>输出：</strong>["Push","Push","Pop","Push"]
<strong>解释：
<strong>解释：
</strong>读取 1 并自动推入数组 -&gt; [1]
读取 2 并自动推入数组，然后删除它 -&gt; [1]
读取 3 并自动推入数组 -&gt; [1,3]
Expand Down Expand Up @@ -81,13 +81,17 @@ tags:

### 方法一：模拟

我们定义 $cur$ 表示当前已经从 `list` 中读取到的数字，初始时 $cur = 0$，用一个数组 $ans$ 存储答案。
我们定义一个变量 $\textit{cur}$ 表示当前待读取的数字，初始时 $\textit{cur} = 1$，用一个数组 $\textit{ans}$ 存储答案。

遍历数组 `target`，对于每个数字 $v$，如果当前要从 `list` 读取的数字小于 $v$，那么我们应该执行 `Push` 和 `Pop` 操作，直到读取的数字等于 $v$，然后执行 `Push` 操作，这样就可以得到数字 $v$。
接下来，我们遍历数组 $\textit{target}$ 中的每个数字 $x$：

遍历结束后，也就构建出了数组 `target`，返回 `ans` 即可。
- 如果 $\textit{cur} < x$，我们将 $\textit{Push}$ 和 $\textit{Pop}$ 依次加入答案，直到 $\textit{cur} = x$；
- 然后我们将 $\textit{Push}$ 加入答案，表示读取数字 $x$；
- 接着，我们将 $\textit{cur}$ 加一，继续处理下一个数字。

时间复杂度 $O(n)$，其中 $n$ 为数组 `target` 的长度。忽略答案的空间消耗，空间复杂度 $O(1)$。
遍历结束后，返回答案数组即可。

时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{target}$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -96,13 +100,14 @@ tags:
```python
class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
cur, ans = 0, []
for v in target:
cur += 1
while cur < v:
ans.extend(['Push', 'Pop'])
ans = []
cur = 1
for x in target:
while cur < x:
ans.extend(["Push", "Pop"])
cur += 1
ans.append('Push')
ans.append("Push")
cur += 1
return ans
```

Expand All @@ -111,14 +116,15 @@ class Solution:
```java
class Solution {
public List<String> buildArray(int[] target, int n) {
int cur = 0;
List<String> ans = new ArrayList<>();
for (int v : target) {
while (++cur < v) {
ans.add("Push");
ans.add("Pop");
int cur = 1;
for (int x : target) {
while (cur < x) {
ans.addAll(List.of("Push", "Pop"));
++cur;
}
ans.add("Push");
++cur;
}
return ans;
}
Expand All @@ -131,14 +137,16 @@ class Solution {
class Solution {
public:
vector<string> buildArray(vector<int>& target, int n) {
int cur = 0;
vector<string> ans;
for (int& v : target) {
while (++cur < v) {
ans.emplace_back("Push");
ans.emplace_back("Pop");
int cur = 1;
for (int x : target) {
while (cur < x) {
ans.push_back("Push");
ans.push_back("Pop");
++cur;
}
ans.emplace_back("Push");
ans.push_back("Push");
++cur;
}
return ans;
}
Expand All @@ -148,32 +156,33 @@ public:
#### Go

```go
func buildArray(target []int, n int) []string {
cur := 0
ans := []string{}
for _, v := range target {
for cur = cur + 1; cur < v; cur++ {
func buildArray(target []int, n int) (ans []string) {
cur := 1
for _, x := range target {
for ; cur < x; cur++ {
ans = append(ans, "Push", "Pop")
}
ans = append(ans, "Push")
cur++
}
return ans
return
}
```

#### TypeScript

```ts
function buildArray(target: number[], n: number): string[] {
const res = [];
let cur = 0;
for (const num of target) {
while (++cur < num) {
res.push('Push', 'Pop');
const ans: string[] = [];
let cur: number = 1;
for (const x of target) {
for (; cur < x; ++cur) {
ans.push('Push', 'Pop');
}
res.push('Push');
ans.push('Push');
++cur;
}
return res;
return ans;
}
```

Expand All @@ -182,18 +191,18 @@ function buildArray(target: number[], n: number): string[] {
```rust
impl Solution {
pub fn build_array(target: Vec<i32>, n: i32) -> Vec<String> {
let mut res = Vec::new();
let mut ans = Vec::new();
let mut cur = 1;
for &num in target.iter() {
while cur < num {
res.push("Push");
res.push("Pop");
for &x in &target {
while cur < x {
ans.push("Push".to_string());
ans.push("Pop".to_string());
cur += 1;
}
res.push("Push");
ans.push("Push".to_string());
cur += 1;
}
res.into_iter().map(String::from).collect()
ans
}
}
```
Expand All @@ -205,21 +214,19 @@ impl Solution {
* Note: The returned array must be malloced, assume caller calls free().
*/
char** buildArray(int* target, int targetSize, int n, int* returnSize) {
char** res = (char**) malloc(sizeof(char*) * n * 2);
char** ans = (char**) malloc(sizeof(char*) * (2 * n));
*returnSize = 0;
int cur = 1;
int i = 0;
for (int j = 0; j < targetSize; j++) {
while (++cur < target[j]) {
res[i] = (char*) malloc(sizeof(char) * 8);
strcpy(res[i++], "Push");
res[i] = (char*) malloc(sizeof(char) * 8);
strcpy(res[i++], "Pop");
for (int i = 0; i < targetSize; i++) {
while (cur < target[i]) {
ans[(*returnSize)++] = "Push";
ans[(*returnSize)++] = "Pop";
cur++;
}
res[i] = (char*) malloc(sizeof(char) * 8);
strcpy(res[i++], "Push");
ans[(*returnSize)++] = "Push";
cur++;
}
*returnSize = i;
return res;
return ans;
}
```

Expand Down
115 changes: 65 additions & 50 deletions solution/1400-1499/1441.Build an Array With Stack Operations/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -93,7 +93,19 @@ The answers that read integer 3 from the stream are not accepted.

<!-- solution:start -->

### Solution 1
### Solution 1: Simulation

We define a variable $\textit{cur}$ to represent the current number to be read, initially set to $\textit{cur} = 1$, and use an array $\textit{ans}$ to store the answer.

Next, we iterate through each number $x$ in the array $\textit{target}$:

- If $\textit{cur} < x$, we add $\textit{Push}$ and $\textit{Pop}$ to the answer alternately until $\textit{cur} = x$;
- Then we add $\textit{Push}$ to the answer, representing reading the number $x$;
- After that, we increment $\textit{cur}$ and continue to process the next number.

After the iteration, we return the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{target}$. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -102,13 +114,14 @@ The answers that read integer 3 from the stream are not accepted.
```python
class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
cur, ans = 0, []
for v in target:
cur += 1
while cur < v:
ans.extend(['Push', 'Pop'])
ans = []
cur = 1
for x in target:
while cur < x:
ans.extend(["Push", "Pop"])
cur += 1
ans.append('Push')
ans.append("Push")
cur += 1
return ans
```

Expand All @@ -117,14 +130,15 @@ class Solution:
```java
class Solution {
public List<String> buildArray(int[] target, int n) {
int cur = 0;
List<String> ans = new ArrayList<>();
for (int v : target) {
while (++cur < v) {
ans.add("Push");
ans.add("Pop");
int cur = 1;
for (int x : target) {
while (cur < x) {
ans.addAll(List.of("Push", "Pop"));
++cur;
}
ans.add("Push");
++cur;
}
return ans;
}
Expand All @@ -137,14 +151,16 @@ class Solution {
class Solution {
public:
vector<string> buildArray(vector<int>& target, int n) {
int cur = 0;
vector<string> ans;
for (int& v : target) {
while (++cur < v) {
ans.emplace_back("Push");
ans.emplace_back("Pop");
int cur = 1;
for (int x : target) {
while (cur < x) {
ans.push_back("Push");
ans.push_back("Pop");
++cur;
}
ans.emplace_back("Push");
ans.push_back("Push");
++cur;
}
return ans;
}
Expand All @@ -154,32 +170,33 @@ public:
#### Go

```go
func buildArray(target []int, n int) []string {
cur := 0
ans := []string{}
for _, v := range target {
for cur = cur + 1; cur < v; cur++ {
func buildArray(target []int, n int) (ans []string) {
cur := 1
for _, x := range target {
for ; cur < x; cur++ {
ans = append(ans, "Push", "Pop")
}
ans = append(ans, "Push")
cur++
}
return ans
return
}
```

#### TypeScript

```ts
function buildArray(target: number[], n: number): string[] {
const res = [];
let cur = 0;
for (const num of target) {
while (++cur < num) {
res.push('Push', 'Pop');
const ans: string[] = [];
let cur: number = 1;
for (const x of target) {
for (; cur < x; ++cur) {
ans.push('Push', 'Pop');
}
res.push('Push');
ans.push('Push');
++cur;
}
return res;
return ans;
}
```

Expand All @@ -188,18 +205,18 @@ function buildArray(target: number[], n: number): string[] {
```rust
impl Solution {
pub fn build_array(target: Vec<i32>, n: i32) -> Vec<String> {
let mut res = Vec::new();
let mut ans = Vec::new();
let mut cur = 1;
for &num in target.iter() {
while cur < num {
res.push("Push");
res.push("Pop");
for &x in &target {
while cur < x {
ans.push("Push".to_string());
ans.push("Pop".to_string());
cur += 1;
}
res.push("Push");
ans.push("Push".to_string());
cur += 1;
}
res.into_iter().map(String::from).collect()
ans
}
}
```
Expand All @@ -211,21 +228,19 @@ impl Solution {
* Note: The returned array must be malloced, assume caller calls free().
*/
char** buildArray(int* target, int targetSize, int n, int* returnSize) {
char** res = (char**) malloc(sizeof(char*) * n * 2);
char** ans = (char**) malloc(sizeof(char*) * (2 * n));
*returnSize = 0;
int cur = 1;
int i = 0;
for (int j = 0; j < targetSize; j++) {
while (++cur < target[j]) {
res[i] = (char*) malloc(sizeof(char) * 8);
strcpy(res[i++], "Push");
res[i] = (char*) malloc(sizeof(char) * 8);
strcpy(res[i++], "Pop");
for (int i = 0; i < targetSize; i++) {
while (cur < target[i]) {
ans[(*returnSize)++] = "Push";
ans[(*returnSize)++] = "Pop";
cur++;
}
res[i] = (char*) malloc(sizeof(char) * 8);
strcpy(res[i++], "Push");
ans[(*returnSize)++] = "Push";
cur++;
}
*returnSize = i;
return res;
return ans;
}
```

Expand Down
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