Longest Common Subsequence (DP)

最新推荐文章于 2021-10-01 11:07:13 发布

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最新推荐文章于 2021-10-01 11:07:13 发布

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原文链接：http://www.cnblogs.com/FLAGyuri/p/5403103.html

本文详细介绍了如何使用动态规划解决最长公共子序列问题，包括状态转移方程、初始化和返回值解释，并通过示例代码展示算法实现过程。

Given two strings, find the longest common subsequence (LCS).

Your code should return the length of LCS.

Example

For "ABCD" and "EDCA", the LCS is "A" (or "D", "C"), return 1.

For "ABCD" and "EACB", the LCS is "AC", return 2.

最长公共子序列的定义：

最长公共子序列问题是在一组序列（通常2个）中找到最长公共子序列（注意：不同于子串，LCS不需要是连续的子串）.

State: f[i][j] 表示在字符串A中前i个字符与B字符串前j个字符的最长LCS。

Fuction： f[i][j] = max(f[i - 1][j], f[i][j - 1]) if (A[i -1] != B[j - 1]) 对应与 “abc” “ab” 和 “ab" 和”abc“。if（A[i - 1] == B[j - 1]) f[i][j] = max(f[i - 1][j], f[i][j - 1], f[i - 1][j -1] + 1).

Initialization: int [][] f = new int[A.length() + 1][B.length() + 1]

Answer:f[A.length()][B.length()]

 1 public class Solution {
 2     /**
 3      * @param A, B: Two strings.
 4      * @return: The length of longest common subsequence of A and B.
 5      */
 6     public int longestCommonSubsequence(String A, String B) {
 7         int m = A.length();
 8         int n = B.length();
 9         if (m == 0 || n == 0) {
10             return 0;
11         }
12         int[][] f = new int[m + 1][n + 1];
13         for (int i = 1; i <= m; i++) {
14             for (int j = 1; j <= n; j++) {
15                 f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
16                 if (A.charAt(i - 1) == B.charAt(j - 1)) {
17                     f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + 1);
18                 }
19             }
20         }
21         return f[m][n];
22     }
23 }