解题思路

每个位置算自己最远能跳到哪里，那么倒过来考虑，更新需要跳到的位置即可

AC代码

class Solution:
    def canJump(self, nums: List[int]) -> bool:
        if len(nums) < 2:
            return True
        nums = [i + v for i, v in enumerate(nums)]

        target = len(nums) - 1
        for i in range(len(nums) - 2, -1, -1):
            if nums[i]  >= target:
                target = i
        
        return target == 0