動機
就是search，不太算是DP? 說到底dp到底是什麼
Problem
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 = 2Output: 3Explanation: 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 = 3Output: 5Explanation: 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 = 0Output: -1Explanation: 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] <= 1090 <= target <= 2 * 109
sol
class Solution:
@cache
def dp(self, i):
if i == len(self.N)-1:
return 0
elif i >= len(self.N):
return -float('inf')
return 1+max([self.dp(j) for j in range(i+1,len(self.N)) if abs(self.N[i]-self.N[j]) <= self.T], default=-float('inf'))
def maximumJumps(self, nums: List[int], target: int) -> int:
self.N = nums
self.T = target
return -1 if self.dp(0) == -float("inf") else self.dp(0)