【LeetCode每天一题】Jump Game II(跳跃游戏II)

Given an array of non-negative integers, you are initially positioned at the first index of the array.Each element in the array represents your maximum jump length at that position.Your goal is to reach the last index in the minimum number of jumps.

Example:

Input: [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.

Note:You can assume that you can always reach the last index.

思路


      这个我们可以参考跳跃游戏I的那个接替思路,只不过这里我们需要增加一个下标变量来记录step数。其余的都和跳跃游戏I一样。时间复杂度为O(n),空间复杂度为O(1)。

图示步骤


    

解决代码


 1 class Solution(object):
 2     def jump(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 7         max_end = 0
 8         end = 0
 9         step = 0            # 记录步数
10         for i in range(len(nums)-1):
11             max_end = max(max_end, nums[i]+i)
12             if i == end:       # 跟新最大下标位置,并步数加一
13                 end = max_end
14                 step += 1
15         return step
原文地址:https://www.cnblogs.com/GoodRnne/p/10734897.html