LeetCode - Jump Game II

You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i] and
• i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

 

Example 1:

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: nums = [2,3,0,1,4]
Output: 2

 

Constraints:

• 1 <= nums.length <= 104
• 0 <= nums[i] <= 1000
• It's guaranteed that you can reach nums[n - 1].

 

My solution:

class Solution(object):
    def jump(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """

        n = len(nums)
        if n == 1:
            return 0
        elif n == 2:
            return 1
        # can jump to end by 1 time
        elif nums[0] >= len(nums) - 1:
            return 1
        # jump the max length
        else:
            max_value = nums[0]
            max_i = 0
            for i in range(1, nums[0] + 1):
                if i + nums[i] > max_value:
                    max_value = i + nums[i]
                    max_i = i
            return 1 + self.jump(nums[max_i:])