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feat: add solutions to lc problem: No.2899 #1809

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acbin merged 1 commit into main from dev
Oct 15, 2023
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100 changes: 97 additions & 3 deletions solution/2800-2899/2899.Last Visited Integers/README.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -54,34 +54,128 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:模拟**

我们直接根据题意模拟即可。在实现上,我们使用一个数组 $nums$ 来存储遍历过的整数,使用一个整数 $k$ 来记录当前连续的 $prev$ 字符串数目。如果当前字符串是 $prev,ドル那么我们就从 $nums$ 中取出第 $|nums| - k$ 个整数,如果不存在,那么就返回 $-1$。

时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $words$ 的长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def lastVisitedIntegers(self, words: List[str]) -> List[int]:
nums = []
ans = []
k = 0
for w in words:
if w == "prev":
k += 1
i = len(nums) - k
ans.append(-1 if i < 0 else nums[i])
else:
k = 0
nums.append(int(w))
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java

class Solution {
public List<Integer> lastVisitedIntegers(List<String> words) {
List<Integer> nums = new ArrayList<>();
List<Integer> ans = new ArrayList<>();
int k = 0;
for (var w : words) {
if ("prev".equals(w)) {
++k;
int i = nums.size() - k;
ans.add(i < 0 ? -1 : nums.get(i));
} else {
k = 0;
nums.add(Integer.valueOf(w));
}
}
return ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
vector<int> lastVisitedIntegers(vector<string>& words) {
vector<int> nums;
vector<int> ans;
int k = 0;
for (auto& w : words) {
if (w == "prev") {
++k;
int i = nums.size() - k;
ans.push_back(i < 0 ? -1 : nums[i]);
} else {
k = 0;
nums.push_back(stoi(w));
}
}
return ans;
}
};
```

### **Go**

```go
func lastVisitedIntegers(words []string) (ans []int) {
nums := []int{}
k := 0
for _, w := range words {
if w == "prev" {
k++
i := len(nums) - k
if i < 0 {
ans = append(ans, -1)
} else {
ans = append(ans, nums[i])
}
} else {
k = 0
x, _ := strconv.Atoi(w)
nums = append(nums, x)
}
}
return
}
```

### **TypeScript**

```ts
function lastVisitedIntegers(words: string[]): number[] {
const nums: number[] = [];
const ans: number[] = [];
let k = 0;
for (const w of words) {
if (w === 'prev') {
++k;
const i = nums.length - k;
ans.push(i < 0 ? -1 : nums[i]);
} else {
k = 0;
nums.push(+w);
}
}
return ans;
}
```

### **...**
Expand Down
100 changes: 97 additions & 3 deletions solution/2800-2899/2899.Last Visited Integers/README_EN.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -48,30 +48,124 @@ For &quot;prev&quot; at index = 4, last visited integer will be 1 as there are a

## Solutions

**Solution 1: Simulation**

We can directly simulate according to the problem statement. In the implementation, we use an array $nums$ to store the traversed integers, and an integer $k$ to record the current number of consecutive $prev$ strings. If the current string is $prev,ドル we take out the $|nums| - k-th$ integer from $nums$. If it does not exist, we return $-1$.

The time complexity is $O(n),ドル where $n$ is the length of the array $words$. The space complexity is $O(n)$.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def lastVisitedIntegers(self, words: List[str]) -> List[int]:
nums = []
ans = []
k = 0
for w in words:
if w == "prev":
k += 1
i = len(nums) - k
ans.append(-1 if i < 0 else nums[i])
else:
k = 0
nums.append(int(w))
return ans
```

### **Java**

```java

class Solution {
public List<Integer> lastVisitedIntegers(List<String> words) {
List<Integer> nums = new ArrayList<>();
List<Integer> ans = new ArrayList<>();
int k = 0;
for (var w : words) {
if ("prev".equals(w)) {
++k;
int i = nums.size() - k;
ans.add(i < 0 ? -1 : nums.get(i));
} else {
k = 0;
nums.add(Integer.valueOf(w));
}
}
return ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
vector<int> lastVisitedIntegers(vector<string>& words) {
vector<int> nums;
vector<int> ans;
int k = 0;
for (auto& w : words) {
if (w == "prev") {
++k;
int i = nums.size() - k;
ans.push_back(i < 0 ? -1 : nums[i]);
} else {
k = 0;
nums.push_back(stoi(w));
}
}
return ans;
}
};
```

### **Go**

```go
func lastVisitedIntegers(words []string) (ans []int) {
nums := []int{}
k := 0
for _, w := range words {
if w == "prev" {
k++
i := len(nums) - k
if i < 0 {
ans = append(ans, -1)
} else {
ans = append(ans, nums[i])
}
} else {
k = 0
x, _ := strconv.Atoi(w)
nums = append(nums, x)
}
}
return
}
```

### **TypeScript**

```ts
function lastVisitedIntegers(words: string[]): number[] {
const nums: number[] = [];
const ans: number[] = [];
let k = 0;
for (const w of words) {
if (w === 'prev') {
++k;
const i = nums.length - k;
ans.push(i < 0 ? -1 : nums[i]);
} else {
k = 0;
nums.push(+w);
}
}
return ans;
}
```

### **...**
Expand Down
19 changes: 19 additions & 0 deletions solution/2800-2899/2899.Last Visited Integers/Solution.cpp
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
class Solution {
public:
vector<int> lastVisitedIntegers(vector<string>& words) {
vector<int> nums;
vector<int> ans;
int k = 0;
for (auto& w : words) {
if (w == "prev") {
++k;
int i = nums.size() - k;
ans.push_back(i < 0 ? -1 : nums[i]);
} else {
k = 0;
nums.push_back(stoi(w));
}
}
return ans;
}
};
20 changes: 20 additions & 0 deletions solution/2800-2899/2899.Last Visited Integers/Solution.go
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
func lastVisitedIntegers(words []string) (ans []int) {
nums := []int{}
k := 0
for _, w := range words {
if w == "prev" {
k++
i := len(nums) - k
if i < 0 {
ans = append(ans, -1)
} else {
ans = append(ans, nums[i])
}
} else {
k = 0
x, _ := strconv.Atoi(w)
nums = append(nums, x)
}
}
return
}
18 changes: 18 additions & 0 deletions solution/2800-2899/2899.Last Visited Integers/Solution.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
class Solution {
public List<Integer> lastVisitedIntegers(List<String> words) {
List<Integer> nums = new ArrayList<>();
List<Integer> ans = new ArrayList<>();
int k = 0;
for (var w : words) {
if ("prev".equals(w)) {
++k;
int i = nums.size() - k;
ans.add(i < 0 ? -1 : nums.get(i));
} else {
k = 0;
nums.add(Integer.valueOf(w));
}
}
return ans;
}
}
14 changes: 14 additions & 0 deletions solution/2800-2899/2899.Last Visited Integers/Solution.py
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
class Solution:
def lastVisitedIntegers(self, words: List[str]) -> List[int]:
nums = []
ans = []
k = 0
for w in words:
if w == "prev":
k += 1
i = len(nums) - k
ans.append(-1 if i < 0 else nums[i])
else:
k = 0
nums.append(int(w))
return ans
16 changes: 16 additions & 0 deletions solution/2800-2899/2899.Last Visited Integers/Solution.ts
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
function lastVisitedIntegers(words: string[]): number[] {
const nums: number[] = [];
const ans: number[] = [];
let k = 0;
for (const w of words) {
if (w === 'prev') {
++k;
const i = nums.length - k;
ans.push(i < 0 ? -1 : nums[i]);
} else {
k = 0;
nums.push(+w);
}
}
return ans;
}

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