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Commit cb9f1f9

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feat: add solutions to lc problem: No.2899 (#1809)
No.2899.Last Visited Integers
1 parent bf5008a commit cb9f1f9

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7 files changed

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lines changed

7 files changed

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‎solution/2800-2899/2899.Last Visited Integers/README.md‎

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@@ -54,34 +54,128 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:模拟**
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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 $-1$。
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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。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def lastVisitedIntegers(self, words: List[str]) -> List[int]:
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nums = []
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ans = []
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k = 0
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for w in words:
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if w == "prev":
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k += 1
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i = len(nums) - k
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ans.append(-1 if i < 0 else nums[i])
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else:
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k = 0
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nums.append(int(w))
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
72-
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class Solution {
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public List<Integer> lastVisitedIntegers(List<String> words) {
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List<Integer> nums = new ArrayList<>();
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List<Integer> ans = new ArrayList<>();
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int k = 0;
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for (var w : words) {
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if ("prev".equals(w)) {
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++k;
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int i = nums.size() - k;
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ans.add(i < 0 ? -1 : nums.get(i));
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} else {
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k = 0;
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nums.add(Integer.valueOf(w));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
78-
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class Solution {
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public:
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vector<int> lastVisitedIntegers(vector<string>& words) {
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vector<int> nums;
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vector<int> ans;
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int k = 0;
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for (auto& w : words) {
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if (w == "prev") {
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++k;
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int i = nums.size() - k;
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ans.push_back(i < 0 ? -1 : nums[i]);
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} else {
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k = 0;
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nums.push_back(stoi(w));
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func lastVisitedIntegers(words []string) (ans []int) {
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nums := []int{}
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k := 0
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for _, w := range words {
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if w == "prev" {
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k++
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i := len(nums) - k
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if i < 0 {
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ans = append(ans, -1)
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} else {
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ans = append(ans, nums[i])
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}
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} else {
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k = 0
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x, _ := strconv.Atoi(w)
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nums = append(nums, x)
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function lastVisitedIntegers(words: string[]): number[] {
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const nums: number[] = [];
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const ans: number[] = [];
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let k = 0;
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for (const w of words) {
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if (w === 'prev') {
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++k;
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const i = nums.length - k;
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ans.push(i < 0 ? -1 : nums[i]);
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} else {
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k = 0;
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nums.push(+w);
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/2800-2899/2899.Last Visited Integers/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -48,30 +48,124 @@ For &quot;prev&quot; at index = 4, last visited integer will be 1 as there are a
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## Solutions
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**Solution 1: Simulation**
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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 $-1$.
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The time complexity is $O(n),ドル where $n$ is the length of the array $words$. The space complexity is $O(n)$.
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<!-- tabs:start -->
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### **Python3**
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```python
56-
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class Solution:
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def lastVisitedIntegers(self, words: List[str]) -> List[int]:
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nums = []
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ans = []
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k = 0
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for w in words:
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if w == "prev":
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k += 1
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i = len(nums) - k
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ans.append(-1 if i < 0 else nums[i])
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else:
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k = 0
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nums.append(int(w))
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return ans
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```
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### **Java**
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```java
62-
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class Solution {
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public List<Integer> lastVisitedIntegers(List<String> words) {
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List<Integer> nums = new ArrayList<>();
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List<Integer> ans = new ArrayList<>();
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int k = 0;
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for (var w : words) {
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if ("prev".equals(w)) {
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++k;
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int i = nums.size() - k;
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ans.add(i < 0 ? -1 : nums.get(i));
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} else {
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k = 0;
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nums.add(Integer.valueOf(w));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
68-
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class Solution {
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public:
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vector<int> lastVisitedIntegers(vector<string>& words) {
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vector<int> nums;
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vector<int> ans;
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int k = 0;
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for (auto& w : words) {
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if (w == "prev") {
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++k;
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int i = nums.size() - k;
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ans.push_back(i < 0 ? -1 : nums[i]);
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} else {
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k = 0;
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nums.push_back(stoi(w));
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}
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}
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return ans;
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}
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};
69123
```
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### **Go**
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```go
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func lastVisitedIntegers(words []string) (ans []int) {
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nums := []int{}
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k := 0
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for _, w := range words {
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if w == "prev" {
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k++
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i := len(nums) - k
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if i < 0 {
136+
ans = append(ans, -1)
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} else {
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ans = append(ans, nums[i])
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}
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} else {
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k = 0
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x, _ := strconv.Atoi(w)
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nums = append(nums, x)
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function lastVisitedIntegers(words: string[]): number[] {
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const nums: number[] = [];
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const ans: number[] = [];
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let k = 0;
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for (const w of words) {
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if (w === 'prev') {
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++k;
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const i = nums.length - k;
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ans.push(i < 0 ? -1 : nums[i]);
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} else {
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k = 0;
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nums.push(+w);
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}
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}
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return ans;
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}
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```
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### **...**
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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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vector<int> lastVisitedIntegers(vector<string>& words) {
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vector<int> nums;
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vector<int> ans;
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int k = 0;
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for (auto& w : words) {
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if (w == "prev") {
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++k;
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int i = nums.size() - k;
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ans.push_back(i < 0 ? -1 : nums[i]);
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} else {
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k = 0;
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nums.push_back(stoi(w));
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}
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
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func lastVisitedIntegers(words []string) (ans []int) {
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nums := []int{}
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k := 0
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for _, w := range words {
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if w == "prev" {
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k++
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i := len(nums) - k
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if i < 0 {
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ans = append(ans, -1)
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} else {
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ans = append(ans, nums[i])
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}
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} else {
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k = 0
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x, _ := strconv.Atoi(w)
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nums = append(nums, x)
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}
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}
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return
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
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class Solution {
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public List<Integer> lastVisitedIntegers(List<String> words) {
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List<Integer> nums = new ArrayList<>();
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List<Integer> ans = new ArrayList<>();
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int k = 0;
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for (var w : words) {
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if ("prev".equals(w)) {
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++k;
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int i = nums.size() - k;
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ans.add(i < 0 ? -1 : nums.get(i));
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} else {
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k = 0;
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nums.add(Integer.valueOf(w));
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}
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}
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return ans;
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}
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,14 @@
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class Solution:
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def lastVisitedIntegers(self, words: List[str]) -> List[int]:
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nums = []
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ans = []
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k = 0
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for w in words:
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if w == "prev":
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k += 1
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i = len(nums) - k
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ans.append(-1 if i < 0 else nums[i])
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else:
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k = 0
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nums.append(int(w))
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return ans
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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function lastVisitedIntegers(words: string[]): number[] {
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const nums: number[] = [];
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const ans: number[] = [];
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let k = 0;
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for (const w of words) {
6+
if (w === 'prev') {
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++k;
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const i = nums.length - k;
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ans.push(i < 0 ? -1 : nums[i]);
10+
} else {
11+
k = 0;
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nums.push(+w);
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}
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}
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return ans;
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}

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