注意「⼦序列」和「⼦串」的区别，⼦串⼀定是连续的，⽽⼦序列不⼀定是连续的。
python
class LengthOfLIS:
"""
300. 最长递增子序列
https://leetcode.cn/problems/longest-increasing-subsequence/description/
"""
def solution(self, nums: List[int]):
"""
方案一: 动态规划
辅助数组 dp，dp[i] 表示以nums[i]结尾的子序列的最大长度
时间复杂度O(N^2)
空间复杂度O(N)
:param nums:
:return:
"""
n = len(nums)
dp = [1] * n
for i in range(1, n):
tmp = 0
for j in range(0, i):
if nums[j] < nums[i]:
tmp = max(tmp, dp[j])
dp[i] = tmp + 1
return max(dp)
def solution2(self, nums: List[int]):
"""
方案二：动态规划
辅助数组 ends，ends[i] 代表目前所有长度为i+1的递增子序列的最小结尾。
可推断出 ends 也是递增序列
时间复杂度O(N*logN)
空间复杂度O(N)
:param nums:
:return:
"""
if not nums:
return 0
n = len(nums)
ends = [0] * n
ends[0] = nums[0]
l, r, m, right = 0, 0, 0, 0
res = 1
for i in range(1, n):
l = 0
r = right
while l <= r:
m = (l + r) // 2
if nums[i] > ends[m]:
l = m + 1
else:
r = m - 1
print(right, l)
right = max(right, l)
ends[l] = nums[i]
res = max(res, l + 1)
return res
def solution3(self, nums: List[int]) -> int:
"""
模拟蜘蛛纸牌
:param nums:
:return:
"""
top = [0] * len(nums)
# 牌堆数初始化为 0
piles = 0
for i in range(len(nums)):
poker = nums[i]
left, right = 0, piles
while left < right:
mid = left + (right - left) // 2
if top[mid] > poker:
right = mid
elif top[mid] < poker:
left = mid + 1
else:
right = mid
# 没找到合适的牌堆，新建⼀堆
if left == piles:
piles += 1
# 把这张牌放到牌堆顶
top[left] = poker
return piles