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.Determine if you are able to reach the last index.
Example 1:
Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximumjump length is 0, which makes it impossible to reach the last index.
思路
这道题解决的办法有很多种。例如DFS，动态规划等，但是感觉这样有些复杂了。有另外一种办法是我们设置一个可以记录下当前下标的范围内达到的最大位置，当遍历过程中的下标大于最大的下标，说明不能达到尾部。因此直接返回结果。当遍历完毕之后说明可以达到最后的位置，返回结果。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 canJump(self, nums):
3 """
4 :type nums: List[int]
5 :rtype: bool
6 """
7 end = 0 # 当前下标所能达到最远的下标
8 max_end = 0
9 for i in range(len(nums)):
10 if end < i: # 说明不能达到尾部
11 return False
12 max_end = max(max_end, nums[i]+i) # 记录达到的最远下标
13
14 if i == end: # 将最远下标重新复制。
15 end = max_end
16
17 return True