思路

1. f[i]表示从0到下标i的最大跳跃次数;
2. f[0]=0, f[i]=f[j]+1, 可以从j跳过来;

class Solution {
public:
    int maximumJumps(vector<int>& nums, int target) {
        int n = nums.size();
        vector<int> f(n, -1);
        f[0] = 0;
        /* 开始递推 */
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (f[j] != -1 && abs(nums[i] - nums[j]) <= target) {
                    f[i] = max(f[i], f[j] + 1);
                }
            }
        }
        return f[n - 1];
    }
};

class Solution:
    def maximumJumps(self, nums: List[int], target: int) -> int:
        n = len(nums)
        f = [-1] * n
        f[0] = 0
        for j in range(1, n):
            for i in range(j):
                if f[i] != -1 and abs(nums[i] - nums[j]) <= target:
                    f[j] = max(f[j], f[i] + 1)
        return f[-1]