1143. 最长公共子序列

• 思路
  • example
  • Longest Common Subsequence (LCS)
  • 双串，二维DP, 不需要连续

dp[i][j]: text1 前i个，text2前j个 结果，i.e., text1: 0,...,i-1; text2: 0,...,j-1
注意对dp[i][j], 对应的最长公共子序列并不一定以i-1th 和 j-1th 结尾。
方便初始化处理, i.e., dp[0][j] = dp[i][0] = 0
Goal: dp[m][n]

• 复杂度. 时间：O(mn), 空间: O(mn)

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        dp = [[0 for _ in range(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][j-1], dp[i-1][j])
        return dp[m][n]

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2) 
        dp = [[0 for _ in range(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[m][n] 

• 可空间优化

1035. 不相交的线

• 思路
  • example
  • 本质就是“最长公共子序列”问题, 不需要连续
• 复杂度. 时间：O(mn), 空间: O(mn)

class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        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[m][n]

class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2) 
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)] 
        dp[0][0] = 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-1][j], dp[i][j-1]) 
        return dp[m][n] 

53. 最大子数组和

• 思路
  • example
  • 连续子数组
    • 滑窗会比较麻烦?
    • DP

dp[i]: 以ith 结尾的最大连续子数组和
目标: max(dp)
if dp[i-1] < 0: dp[i] = nums[i]
else: dp[i] = dp[i-1] + nums[i]

• 复杂度. 时间：O(n), 空间: O(n)

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

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

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

小结

• 1维，2维，？
• 前i个 或者 以ith结尾，？
• 目标

583. 两个字符串的删除操作

• 思路
  • example

双串, dp[i][j]
dp[i][j]: word1[0,...,i-1], word2[0,...,j-1]. 前i个，前j个

• 复杂度. 时间：O(mn), 空间: O(mn)

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        for j in range(n+1): # !!!
            dp[0][j] = j
        for i in range(m+1): # !!!
            dp[i][0] = i   
        for i in range(1, m+1):
            for j in range(1, n+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1
        return dp[m][n]  

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        for j in range(1, n+1):
            dp[0][j] = j 
        for i in range(1, m+1):
            dp[i][0] = i
        for i in range(1, m+1):
            for j in range(1, n+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1 
        return dp[m][n] 

712. 两个字符串的最小ASCII删除和

• 思路
  • example

双串
ASCII: ord('a')

• 复杂度. 时间：O(mn), 空间: O(mn)

class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        m, n = len(s1), len(s2)
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        for j in range(1, n+1):
            dp[0][j] = dp[0][j-1] + ord(s2[j-1])
        for i in range(1, m+1):
            dp[i][0] = dp[i-1][0] + ord(s1[i-1])
        for i in range(1, m+1):
            for j in range(1, n+1):
                if s1[i-1] == s2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j]+ord(s1[i-1]), dp[i][j-1]+ord(s2[j-1]))
        return dp[m][n]

class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        m, n = len(s1), len(s2) 
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        for j in range(1, n+1):
            dp[0][j] = dp[0][j-1] + ord(s2[j-1])
        for i in range(1, m+1):
            dp[i][0] = dp[i-1][0] + ord(s1[i-1])
        for i in range(1, m+1):
            for j in range(1, n+1):
                if s1[i-1] == s2[j-1]:
                    dp[i][j] = dp[i-1][j-1] 
                else:
                    dp[i][j] = min(dp[i-1][j]+ord(s1[i-1]), dp[i][j-1]+ord(s2[j-1]))
        return dp[m][n]