1.最长上升子序列
输入： nums = [10,9,2,5,3,7,101,18]
输出： 4
解释： 最长递增子序列是 [2,3,7,101]，因此长度为 4 。
nums = [10, 9, 2, 5, 3, 7, 101, 18]
if not nums:
print(0)
dp = [1 for _ in range(len(nums))]
for i in range(1, len(nums)):
for j in range(i):
if nums[j] < nums[i]:
dp[i] = max(dp[i], dp[j] + 1)
print(max(dp))
2.最长--连续--子序列
输入： nums = [10,9,2,5,3,7,101,18]
输出： 3
解释： 最长递增子序列是 [3,7,101]，因此长度为 3 。
nums = [10, 9, 2, 5, 3, 7, 101, 18]
dp = [1 for _ in range(len(nums))]
for i in range(2, len(nums)):
if nums[i] >= nums[i - 1]:
dp[i] = dp[i - 1] + 1
else:
dp[i] = 1
print(max(dp))
3.最长公共子序列
基础版
输入：text1 = "abcde", text2 = "ace"
输出：3
解释：最长公共子序列是 "ace" ，它的长度为 3 。
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
res = ""
# 记录最长的公共子序列
if m * n == 0:
return 0
dp = [[0] * (n + 1) for _ in range(m + 1)]
# dp[i][j]表示text1[:i]与text2[:j]的最长公共子序列长度
for i in range(1, m + 1):
for j in range(1, n + 1):
if text1[i - 1] == text2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
res += text1[i - 1]
else:
dp[i][j] = max([dp[i - 1][j], dp[i][j - 1]])
print(res)
return dp[m][n]
text1 = "abcde"
text2 = "ace"
obj = Solution()
res = obj.longestCommonSubsequence(text1, text2)
print(res)
简易版
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
n=len(text1)
m=len(text2)
@cache
def dfs(i,j):
if i<0 or j<0:
return 0
if text1[i]==text2[j]:
return dfs(i-1,j-1)+1 #+1意味着text1[1]=text2[j]都选
return max(dfs(i-1,j),dfs(i,j-1)) #都不选
return dfs(n-1,m-1)
4.最长公共子串
注意最长公共子序列和最长公共子串的区别
最长公共子串要求连续
输入：text1 = "helloworld", text2 = "loop"
输出：2
解释：最长公共子序列是 "lo" ，它的长度为 2 。
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
res = ""
# 记录最长的公共子序列
if m * n == 0:
return 0
dp = [[0] * (n + 1) for _ in range(m + 1)]
# dp[i][j]表示text1[:i]与text2[:j]的最长公共子序列长度
result = 0
for i in range(1, m + 1):
for j in range(1, n + 1):
if text1[i - 1] == text2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
res += text1[i - 1]
result = max(result, dp[i][j])
print(result)
return dp[m][n]
text1 = "helloworld"
text2 = "loop"
obj = Solution()
res = obj.longestCommonSubsequence(text1, text2)
