代码随想录算法训练营第47天 | 动态序列11：序列专题2

代码随想录算法训练营第天 |

1143.最长公共子序列
https://leetcode.cn/problems/longest-common-subsequence/description/
代码随想录
https://programmercarl.com/1143.最长公共子序列.html#算法公开课
1035.不相交的线
https://leetcode.cn/problems/uncrossed-lines/description/
代码随想录
https://programmercarl.com/1035.不相交的线.html#其他语言版本
53.最大子序和
https://leetcode.cn/problems/maximum-subarray/description/
代码随想录
https://programmercarl.com/0053.最大子序和（动态规划）.html#其他语言版本
392.判断子序列
https://leetcode.cn/problems/is-subsequence/submissions/549842019/
代码随想录
https://programmercarl.com/0392.判断子序列.html#算法公开课

1143.最长公共子序列

题解

• dp[i][j]：第i-1个text1和j-1个text2字符串的最长公共子序列；
• 初始化 dp = m+1 * n+1 的0矩阵；
• 递推：
  • 字符串一致时：
    dp[i][j] = dp[i-1][j-1]+1
  • 字符串不一致时：
    dp[i][j] = max(dp[i-1][j],dp[i][j-1])

点击查看代码

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m = len(text1)
        n = len(text2)

        dp = [[0]*(n+1) for _ in range(m+1)]

        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
                else:
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1])
        return dp[-1][-1]

1035. 不相交的线

题解

• 从逻辑上来讲和最长公共子序列一致
• 直接套的结果是一样的

点击查看代码

class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        m = len(nums1)
        n = len(nums2)

        dp = [[0]*(n+1) for _ in range(m+1)]
        res = 0
        for i in range(1,m+1):
            for j in range(1,n+1):
                if nums1[i-1]==nums2[j-1]:
                    dp[i][j] = dp[i-1][j-1]+1
                else:
                    dp[i][j] = max(dp[i][j-1],dp[i-1][j])
        return dp[-1][-1]

53. 最大子数组和

• dp：第i个字符串之前最大的连续和；
• 初始化：最小是第一个数；dp[0] = nums[0]; res = nums[0]
• 递推：
  • dp[i] = max(dp[i-1]+nums[i],nums[i])

点击查看代码

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        dp = [0]*len(nums)
        dp[0] = nums[0]
        res = dp[0]
        for i in range(1,len(nums)):
            dp[i] = max(dp[i-1]+nums[i],nums[i])
            if dp[i]>res:
                res = dp[i]
        return res

392.判断子序列

题解代码

• dp[i][j]:第i-1个s字符串和j-1个t字符串的相同字符串长度；
• 初始化：为了不用初始化第一行、第一列：dp[i][j]
• 递推：
  • 如果s[i-1] == t[j-1] dp[i][j]=dp[i-1][j-1]+1
  • 如果s[i-1]!=t[j-1] dp[i][j] = dp[i][j-1] 跳过t的这个点即可
• 如果dp[i][j]长度和s长度一样 说明符合要求；

点击查看代码

import numpy as np
class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        m = len(s)
        n = len(t)

        dp = [[0] * (n+1) for _ in range(m+1)] 


        for i in range(1,m+1):
            for j in range(1,n+1):
                if s[i-1] == t[j-1]:
                    dp[i][j] = dp[i-1][j-1]+1
                else:
                    dp[i][j] = dp[i][j-1]
                if dp[i][j]>=m:
                    return True
        if dp[-1][-1]>=m:
            return True
        else:
            return False