Day 54 DP：子序列问题 小结

子序列问题

• 一般指“不连续的序列”
• 1个序列，2个序列，。。。
• 最长，最少，。。。
• DP: 复杂度一般是 或
• 经典题目
  • LIS 单串
  • LCS 双串
  • 编辑距离

300. 最长递增子序列

• 子序列：不需要连续
• 单串DP

dp[i]: 以ith结尾

• 复杂度. 时间：, 空间:

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

dp[i]: 长度为i的LIS的最后一个元素的最小可能值。这样就能构造尽可能长的递增子序列。
初始化dp[0] = float('-inf')
dp是严格递增的 （单调栈?)

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        def BinarySearch_insert(dp, target):
            left, right = 0, len(dp)-1
            while left <= right:
                mid = left + (right - left) // 2
                if dp[mid] > target:
                    right = mid - 1
                elif dp[mid] < target:
                    left = mid + 1
                else: # nums[mid] == target
                    right = mid - 1
            return left  # it's possible to have left = len(dp) + 1
        n = len(nums)
        dp = [float('-inf')]
        res = 0
        for i in range(n):
            # 在dp中找最小的index使得dp[index] >= nums[i], 二分找左边界
            insert_idx = BinarySearch_insert(dp, nums[i])
            if insert_idx == len(dp):
                dp.append(nums[i])
                res += 1
            else:
                dp[insert_idx] = nums[i]
        return res

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        def BinarySearch(nums, target):
            n = len(nums)
            left, right = 0, n-1
            while left <= right:
                mid = left + (right - left) // 2
                if nums[mid] < target:
                    left = mid + 1
                else:
                    right = mid - 1
            return left 
        n = len(nums)
        stack = [nums[0]]
        for i in range(1, n):
            idx = BinarySearch(stack, nums[i]) 
            if idx == len(stack):
                stack.append(nums[i]) 
            else:
                stack[idx] = nums[i] 
        return len(stack) 

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] * (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]

72. 编辑距离

dp[i][j]

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][j-1], dp[i-1][j], dp[i-1][j-1]) + 1
        return dp[m][n] 

其它类似问题

583. 两个字符串的删除操作

• 思路
  • example
  • 最小步数

dp[i][j]: word1 前i个，word2前j个 （word1: 0, ..., i-1; word2: 0, ..., j-1)

• 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] * (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] # 不需删除
                else:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1 # 删除word1的i-1th或word2的j-1th
        return dp[m][n]

712. 两个字符串的最小ASCII删除和

• 思路
  • example
  • 类似上题
• 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] * (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] # 不需删除
                else:
                    dp[i][j] = min(dp[i-1][j] + ord(s1[i-1]), dp[i][j-1] + ord(s2[j-1])) # 删除s1的i-1th或s2的j-1th
        return dp[m][n]

1312. 让字符串成为回文串的最少插入次数

• 单串，区间DP (二维）

dp[i][j]: 使 s[i], ..., s[j] 成为回文串的 最少插入次数
注意遍历顺序 （根据递推公式）

class Solution:
    def minInsertions(self, s: str) -> int:
        n = len(s)  
        dp = [[0] * n for _ in range(n)]
        for i in range(n-1, -1, -1):
            for j in range(i, n):
                if j == i:
                    dp[i][j] = 0 # 单个字符已经是回文
                else: # i < j
                    if s[i] == s[j]:
                        dp[i][j] = dp[i+1][j-1]
                    else: # s[i] != s[j]
                        dp[i][j] = min(dp[i+1][j], dp[i][j-1]) + 1
                        #              j+1插入s[i]  i-1插入s[j]
        return dp[0][n-1]

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

516. 最长回文子序列

• 不连续子序列
• 单串，区间DP (二维）

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n = len(s) 
        dp = [[0]*n for _ in range(n)]
        for i in range(n-1, -1, -1):
            for j in range(i, n):
                if i == j:
                    dp[i][j] = 1
                else: # i < j
                    if s[i] == s[j]:
                        dp[i][j] = dp[i+1][j-1] + 2
                    else: # s[i] != s[j]
                        dp[i][j] = max(dp[i+1][j], dp[i][j-1]) # 最长
        return dp[0][n-1]