题目
Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
Example 1:
Input: nums = [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.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
1 <= nums.length <= 2500-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?
思路
- 动态规划
- 耐心排序
代码
python版本：
class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
dp = [0]*len(nums)
for i in range(len(nums)):
less = [dp[j] for j in range(i) if nums[j] < nums[i]]
dp[i] = max(less)+1 if less else 1
return max(dp)
# 耐心排序
class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
helper = []
for n in nums:
if not helper:
helper.append([n])
elif n<=helper[0][-1]:
helper[0][-1]=n
elif n>helper[-1][-1]:
helper.append(helper[-1]+[n])
else:
l,r=0,len(helper)
while l<r:
m=(l+r)//2
if helper[m][-1]<n:
l=m+1
else:
r=m
helper[l][-1]=n
return len(helper[-1])
