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View Code? Open in Web Editor NEW算法精粹--举一反三,抛弃题海战术
License: GNU General Public License v3.0
算法精粹--举一反三,抛弃题海战术
License: GNU General Public License v3.0
首先,谢谢提供如此好的一个教程帮助初学者或像我这种忘记的差不多的人重新学习。
示例代码似乎有误,我在leecode中尝试了,不能通过,因为nums中的元素被替换,所以好比[1,1,1,2,2,3]这样的list会出错。
我找到一个稍微简单的办法处理这种问题,直接删除list中重复的元素,示例代码如下(n=2,使用其他n>2的情况):
class Solution(object):
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) <= 2:
return len(nums)
idx = 2
while idx < len(nums):
if nums[idx-2]== nums[idx]:
del nums[idx]
else:
idx+=1
return len(nums)
解法1的状态转移方程应该是:
f[j] = max{f[i], 0 <= i < j && nums[i] < nums[j]} + 1
原来的写成了:
f[j] = max{f[i], 0 <= i < j && f[i] < f[j]} + 1
Remove Duplicates from Sorted Array II 代码 2 注解:
下面是一个更简洁的版本。上面的代码略长,不过扩展性好一些,例如将
occur < 2
改为occur < 3
,就变成了允许重复最多 3 次。
代码 1 所用变量名为N = 2
,而非occur < 2
他的原理是什么,有没有好心人讲解一下
Remove Duplicates from Sorted Array II 代码 2 注解:
下面是一个更简洁的版本。上面的代码略长,不过扩展性好一些,例如将
occur < 2
改为occur < 3
,就变成了允许重复最多 3 次。
代码 1 所用变量名为N = 2
,而非occur < 2
如题
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