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feat: add solutions to lc problem: No.1696
No.1696.Jump Game VI
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‎solution/1600-1699/1696.Jump Game VI/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划 + 单调队列优化**
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我们定义 $f[i]$ 表示到达下标 $i$ 的最大得分，那么 $f[i]$ 的值可以从 $f[j]$ 转移而来，其中 $j$ 满足 $i - k \leq j \leq i - 1$。因此我们可以使用动态规划求解。
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状态转移方程为：
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$$
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f[i] = \max_{j \in [i - k, i - 1]} f[j] + nums[i]
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$$
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我们可以使用单调队列优化状态转移方程，具体做法是维护一个单调递减的队列，队列中存储的是下标 $j$，并且队列中的下标对应的 $f[j]$ 值是单调递减的。在进行状态转移时，我们只需要取出队首的下标 $j$，即可得到 $f[j]$ 的最大值，然后将 $f[i]$ 的值更新为 $f[j] + nums[i]$ 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。
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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 maxResult(self, nums: List[int], k: int) -> int:
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n = len(nums)
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f = [0] * n
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q = deque([0])
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for i in range(n):
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if i - q[0] > k:
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q.popleft()
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f[i] = nums[i] + f[q[0]]
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while q and f[q[-1]] <= f[i]:
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q.pop()
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q.append(i)
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return f[-1]
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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
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class Solution {
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public int maxResult(int[] nums, int k) {
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int n = nums.length;
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int[] f = new int[n];
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(0);
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for (int i = 0; i < n; ++i) {
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if (i - q.peekFirst() > k) {
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q.pollFirst();
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}
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f[i] = nums[i] + f[q.peekFirst()];
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while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
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q.pollLast();
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}
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q.offerLast(i);
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}
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return f[n - 1];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxResult(vector<int>& nums, int k) {
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int n = nums.size();
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int f[n];
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f[0] = 0;
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deque<int> q = {0};
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for (int i = 0; i < n; ++i) {
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if (i - q.front() > k) q.pop_front();
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f[i] = nums[i] + f[q.front()];
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while (!q.empty() && f[q.back()] <= f[i]) q.pop_back();
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q.push_back(i);
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}
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return f[n - 1];
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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 maxResult(nums []int, k int) int {
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n := len(nums)
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f := make([]int, n)
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q := []int{0}
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for i, v := range nums {
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if i-q[0] > k {
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q = q[1:]
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}
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f[i] = v + f[q[0]]
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for len(q) > 0 && f[q[len(q)-1]] <= f[i] {
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q = q[:len(q)-1]
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}
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q = append(q, i)
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}
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return f[n-1]
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}
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```
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### **...**

‎solution/1600-1699/1696.Jump Game VI/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maxResult(self, nums: List[int], k: int) -> int:
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n = len(nums)
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f = [0] * n
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q = deque([0])
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for i in range(n):
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if i - q[0] > k:
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q.popleft()
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f[i] = nums[i] + f[q[0]]
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while q and f[q[-1]] <= f[i]:
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q.pop()
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q.append(i)
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return f[-1]
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```
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### **Java**
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```java
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class Solution {
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public int maxResult(int[] nums, int k) {
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int n = nums.length;
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int[] f = new int[n];
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(0);
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for (int i = 0; i < n; ++i) {
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if (i - q.peekFirst() > k) {
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q.pollFirst();
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}
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f[i] = nums[i] + f[q.peekFirst()];
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while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
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q.pollLast();
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}
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q.offerLast(i);
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}
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return f[n - 1];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxResult(vector<int>& nums, int k) {
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int n = nums.size();
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int f[n];
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f[0] = 0;
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deque<int> q = {0};
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for (int i = 0; i < n; ++i) {
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if (i - q.front() > k) q.pop_front();
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f[i] = nums[i] + f[q.front()];
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while (!q.empty() && f[q.back()] <= f[i]) q.pop_back();
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q.push_back(i);
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}
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return f[n - 1];
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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 maxResult(nums []int, k int) int {
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n := len(nums)
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f := make([]int, n)
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q := []int{0}
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for i, v := range nums {
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if i-q[0] > k {
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q = q[1:]
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}
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f[i] = v + f[q[0]]
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for len(q) > 0 && f[q[len(q)-1]] <= f[i] {
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q = q[:len(q)-1]
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}
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q = append(q, i)
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}
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return f[n-1]
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}
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```
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### **...**

‎solution/1600-1699/1696.Jump Game VI/Solution.cpp

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class Solution {
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public:
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int maxResult(vector<int>& nums, int k) {
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int n = nums.size();
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int f[n];
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f[0] = 0;
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deque<int> q = {0};
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for (int i = 0; i < n; ++i) {
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if (i - q.front() > k) q.pop_front();
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f[i] = nums[i] + f[q.front()];
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while (!q.empty() && f[q.back()] <= f[i]) q.pop_back();
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q.push_back(i);
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}
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return f[n - 1];
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}
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};

‎solution/1600-1699/1696.Jump Game VI/Solution.go

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func maxResult(nums []int, k int) int {
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n := len(nums)
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f := make([]int, n)
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q := []int{0}
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for i, v := range nums {
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if i-q[0] > k {
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q = q[1:]
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}
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f[i] = v + f[q[0]]
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for len(q) > 0 && f[q[len(q)-1]] <= f[i] {
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q = q[:len(q)-1]
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}
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q = append(q, i)
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}
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return f[n-1]
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}

‎solution/1600-1699/1696.Jump Game VI/Solution.java

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class Solution {
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public int maxResult(int[] nums, int k) {
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int n = nums.length;
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int[] f = new int[n];
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(0);
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for (int i = 0; i < n; ++i) {
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if (i - q.peekFirst() > k) {
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q.pollFirst();
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}
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f[i] = nums[i] + f[q.peekFirst()];
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while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
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q.pollLast();
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}
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q.offerLast(i);
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}
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return f[n - 1];
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}
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}

‎solution/1600-1699/1696.Jump Game VI/Solution.py

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class Solution:
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def maxResult(self, nums: List[int], k: int) -> int:
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n = len(nums)
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f = [0] * n
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q = deque([0])
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for i in range(n):
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if i - q[0] > k:
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q.popleft()
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f[i] = nums[i] + f[q[0]]
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while q and f[q[-1]] <= f[i]:
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q.pop()
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q.append(i)
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return f[-1]

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