LeetCode #4 Longest Common Subsequence

喜欢ヅ旅行 2023-08-17 17:47 105阅读 0赞

### Question ###

Given two strings `text1` and `text2`, return the length of their longest common subsequence.

A *subsequence* of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A *common subsequence* of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

动归经典问题，最长公共子序列（LCS），注意是子序列而不是子串

Example 1:

Input: text1 = "abcde", text2 = "ace" 
    Output: 3  
    Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
    Output: 3
    Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
    Output: 0
    Explanation: There is no such common subsequence, so the result is 0.

### 动态规划法 ###

先确定状态（或最优子结构），c(i, j)表示str1中前i个字符和str2中前j个字符中的最长公共子序列的长度

可以举一个例子思考来写出状态转移方程：

当 str1\[i\] == str2\[j\]，c(i, j) = c(i-1, j-1) + 1

当 str1\[i\] != str2\[j\]，c(i, j) = max(c(i, j-1), c(i-1, j))

当 i == 0 **或** j == 0 时，若 str1\[i\] == str2\[j\]，c(i, j) = 1；str1\[i\] != str2\[j\]，c(i, j) = 0  (边界条件）

（这个是“或”而不是“和”，非常重要，第一次写的时候就搞错了）

以 str1 = "ABCBDAB", str2 = "BDCABA" 为例来说明枚举的顺序

![1404499-20190922083348519-566803541.png][]

箭头表示“source”，因此填表顺序可以为从上到下，从左到右

从图中可知，边界条件并不只是左上角，而是最上面一条边和最左边一条边

（其实我的算法和这个图是有区别的，这个图将index=0表示为空串了，而我的算法将index=0表示第一个字符）

class Solution:
        def longestCommonSubsequence(self, text1: str, text2: str) -> int:
            len_1 = len(text1)
            len_2 = len(text2)
            dp = [[0 for j in text2] for i in text1]
            for i in range(len_1):
                for j in range(len_2):
                    if (i == 0 or j == 0) and text1[i] == text2[j]:
                        dp[i][j] = 1 
                    elif text1[i] == text2[j]:
                        dp[i][j] = dp[i-1][j-1] + 1
                    else:
                        dp[i][j] = max(dp[i][j-1], dp[i-1][j])
                    # print(dp[i][j], text1[i], text2[j])
            return dp[len_1-1][len_2-1]

时间复杂度为O(n\*m)

打印路径：

定义一个全局变量flag\[i\]\[j\]用于记录每次的路径是“source”来自左上角/左边/上边，再用一个输出函数递归地打印出来（可参考下方链接）

参考：

[https://www.cnblogs.com/wkfvawl/p/9362287.html][https_www.cnblogs.com_wkfvawl_p_9362287.html] （以该题为例从某角度理解动态规划）

转载于:https://www.cnblogs.com/sbj123456789/p/11565809.html

[1404499-20190922083348519-566803541.png]: /images/20230808/071f8670412648069220b1a7bc61fd2c.png
[https_www.cnblogs.com_wkfvawl_p_9362287.html]: https://www.cnblogs.com/wkfvawl/p/9362287.html