class Solution(object):
# 思路:
# 维护一个reach变量,代码能走到的最远index
# while循环, i <= reach,
# 每次更新reach
def canJump(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
reach = 0
i = 0
while i <= reach:
reach = max(reach, i+nums[i])
if reach >= len(nums) - 1:
return True
i += 1
return False
前篇 55. Jump Game 45. Jump Game II 感谢Jason_Yuan大神优秀的解析：链接 ...
55. Jump Game Every time we try the furthest position we ...
55. Jump Game Given an array of non-negative integers, yo...
55. 跳跃游戏[https://leetcode-cn.com/problems/jump-game/] 动态规...
55. 跳跃游戏 题目来源：https://leetcode-cn.com/problems/jump-game ...
55. 跳跃游戏[https://leetcode-cn.com/problems/jump-game/] 方法一...
55.跳跃游戏 https://leetcode-cn.com/problems/jump-game/[https...
分类：Greedy 时间复杂度: O(n) 55. Jump Game Given an array of non...
题目分析 纪念一下，解出来的第一道击败100%其他答案的题目，给自己鼓个掌，继续努力 代码 时间复杂度为O(n)，...
题目 Given an array of non-negative integers, you are initi...
本文标题：55. Jump Game
本文链接：https://www.haomeiwen.com/subject/iksfrctx.html
网友评论