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longest-increasing-subsequence.py
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58 lines (39 loc) · 1.53 KB
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'''
Given an unsorted array of integers, find the length of longest increasing subsequence.
Example:
Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Note:
There may be more than one LIS combination, it is only necessary for you to return the length.
Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?
'''
class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
# Approach two O(nlogn)
# dp 列表记录上升序列,可能有多个,我们记录最后一个数最小的上升序列。
# 每次遍历dp,找出第一个比当前数字大的数字进行替换(二分查找)
if not nums : return 0
res = [nums[0]]
for i in range(1,len(nums)):
if nums[i] > res[-1]:
res.append(nums[i])
else:
l , r = 0 , len(res)-1
while l < r :
mid = l + (r - l) // 2
if nums[i] <= res[mid]:
r = mid
else:
l = mid + 1
res[l] = nums[i]
return len(res)
# Approach one 人人为我
# if not nums : return 0
# res = [1 for _ in range(len(nums))]
# for i in range(1,len(nums)):
# for j in range(0,i):
# if nums[j] < nums[i]:
# res[i] = max(res[i], res[j]+1)
# return max(res)