72. Edit Distance

https://leetcode.com/problems/edit-distance/

solution

• 动态规划

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m = len(word1)
        n = len(word2)

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

        for i in range(m+1):
            dp[i][0] = i
       
        for j in range(n+1):
            dp[0][j] = j
        
        for i in range(1, m+1):  # 注意递推范围
            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-1], dp[i-1][j], dp[i][j-1]) + 1
        return dp[-1][-1]

时间复杂度：O() 空间复杂度：O()

• dfs

时间复杂度：O() 空间复杂度：O()

follow up

392.判断子序列

• 类似只允许删除的编辑距离，注意dp含义不是原本要求的定性是否问题，而是转化为定量多少问题

• 同时dp含义还要融入，最长重复子串探讨过的dp含义，以-结尾的序列，还是截至到-的序列

# dp[i][j]: 以下标i-1为结尾的字符串s，和以下标j-1为结尾的字符串t，相同子序列的长度为dp[i][j]
class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        dp = [[0] * (len(t)+1) for _ in range(len(s)+1)]
        for i in range(1, len(s)+1):
            for j in range(1, len(t)+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[-1][-1] == len(s):
            return True
        return False

双指针

def isSubsequence(self, s: str, t: str) -> bool:
    i, j  = 0, 0

    while i < len(s) and j < len(t):
        if s[i] == t[j]:
            i += 1
        j += 1
    
    return i == len(s)