达到末尾下标所需的最大跳跃次数

You are given a 0-indexed array nums of n integers and an integer target.

You are initially positioned at index 0. In one step, you can jump from index i to any index j such that:

• 0 <= i < j < n
• -target <= nums[j] - nums[i] <= target

Return the maximum number of jumps you can make to reach index n - 1.

If there is no way to reach index n - 1, return -1.

 

Example 1:

Input: nums = [1,3,6,4,1,2], target = 2
Output: 3
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1. 
- Jump from index 1 to index 3.
- Jump from index 3 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 3 jumps. Hence, the answer is 3. 

Example 2:

Input: nums = [1,3,6,4,1,2], target = 3
Output: 5
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1.
- Jump from index 1 to index 2.
- Jump from index 2 to index 3.
- Jump from index 3 to index 4.
- Jump from index 4 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 5 jumps. Hence, the answer is 5. 

Example 3:

Input: nums = [1,3,6,4,1,2], target = 0
Output: -1
Explanation: It can be proven that there is no jumping sequence that goes from 0 to n - 1. Hence, the answer is -1. 

 

Constraints:

• 2 <= nums.length == n <= 1000
• -109 <= nums[i] <= 109
• 0 <= target <= 2 * 109

/*
 * 思路：
 * 使用动态规划和Java Stream API来简化部分操作。
 * 和上面的思路类似，维护一个dp数组来存储到达每个位置的最大跳跃次数。
 * 使用IntStream遍历来实现动态规划的跳跃过程。
 */

import java.util.*;
import java.util.stream.*;

public class Solution {
    public int maxJump(int[] nums, int target) {
        int n = nums.length;
        int[] dp = new int[n];
        Arrays.fill(dp, -1);
        dp[0] = 0;

        IntStream.range(0, n).forEach(i -> {
            if (dp[i] != -1) {
                IntStream.range(i + 1, n).forEach(j -> {
                    if (nums[j] - nums[i] >= -target && nums[j] - nums[i] <= target) {
                        dp[j] = Math.max(dp[j], dp[i] + 1);
                    }
                });
            }
        });

        return dp[n - 1];
    }
}

该题解采用了线段树（ZKW线段树的非递归版本）来求解问题。首先，为了应对题目中允许的跳跃范围 [-target, target]，题解将数组中的每个元素 x 扩展到三个可能的跳跃值 x-target, x, x+target，并去重排序。这种扩展允许在 O(log n) 时间内查询和更新任何区间的最大值。通过映射每个扩展值到其索引，可以用线段树高效查询和更新最大的跳跃次数。遍历 nums 数组，对于每个元素 x，查询在 [x-target, x+target] 范围内的最大跳跃次数，并在线段树中更新当前 x 的位置。最后，检查是否能到达 n-1 索引处，并返回相应的跳跃次数。如果不能到达，则返回 -1。

O(n log n)

O(n)