leetcode-mid-dynamic programming-55. Jump Game
mycode 71.47%
思路:
既然要到达终点,那么俺就可以倒推,要想到达n,可以有以下情况
1)到达n-1,然后该位置最少可以走一步
2)到达n-2,然后该位置最少可以走两步
3)到达n-3,然后该位置最少可以走三步
。。。。
emmmm...本来想按照这个方法写个函数,发现跑不通。。。。我就干脆先把list倒过来,当下个数大于等于,他能到达上一个数,但是当这个数暂时不能到达时,我就需要给下一个数加一个step的要求啦
class Solution(object):
def canJump(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
nums[:] = nums[::-1]
length = len(nums)
step = 1
for i in range(1,length):
if nums[i] >= step:
step = 1
continue
else:
step += 1
return step == 1
参考:
跟我的差不多,区别:我的目标--不断衡量距离,他的目标--不断改变目的地
class Solution(object):
def canJump(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
if not nums or len(nums)==1 : return True
des = len(nums) - 1
for i in range(len(nums)-2,-1,-1):
if nums[i] >= des - i:
des = i
return des == 0
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