# Longest Common Subsequence- tags: [[DP_Two_Sequence](# "一般有两个数组或者两个字符串，计算其匹配关系. 通常可用 `f[i][j]`表示第一个数组的前 i 位和第二个数组的前 j 位的关系。")]### Source- lintcode: [(77) Longest Common Subsequence](http://www.lintcode.com/en/problem/longest-common-subsequence/)~~~Given two strings, find the longest common subsequence (LCS).Your code should return the length of LCS.Have you met this question in a real interview? YesExampleFor "ABCD" and "EDCA", the LCS is "A" (or "D", "C"), return 1.For "ABCD" and "EACB", the LCS is "AC", return 2.ClarificationWhat's the definition of Longest Common Subsequence?https://en.wikipedia.org/wiki/Longest_common_subsequence_problemhttp://baike.baidu.com/view/2020307.htm~~~### 题解求最长公共子序列的数目，注意这里的子序列可以不是连续序列，务必问清楚题意。求『最长』类的题目往往与动态规划有点关系，这里是两个字符串，故应为双序列动态规划。这道题的状态很容易找，不妨先试试以`f[i][j]`表示字符串 A 的前 `i` 位和字符串 B 的前 `j` 位的最长公共子序列数目，那么接下来试试寻找其状态转移方程。从实际例子`ABCD`和`EDCA`出发，首先初始化`f`的长度为字符串长度加1，那么有`f[0][0] = 0`, `f[0][*] = 0`, `f[*][0] = 0`, 最后应该返回`f[lenA][lenB]`. 即 f 中索引与字符串索引对应(字符串索引从1开始算起)，那么在A 的第一个字符与 B 的第一个字符相等时，`f[1][1] = 1 + f[0][0]`, 否则`f[1][1] = max(f[0][1], f[1][0])`。推而广之，也就意味着若`A[i] == B[j]`, 则分别去掉这两个字符后，原 LCS 数目减一，那为什么一定是1而不是0或者2呢？因为不管公共子序列是以哪个字符结尾，在`A[i] == B[j]`时 LCS 最多只能增加1. 而在`A[i] != B[j]`时，由于`A[i]` 或者 `B[j]` 不可能同时出现在最终的 LCS 中，故这个问题可进一步缩小，`f[i][j] = max(f[i - 1][j], f[i][j - 1])`. 需要注意的是这种状态转移方程只依赖最终的 LCS 数目，而不依赖于公共子序列到底是以第几个索引结束。### Python~~~class Solution: """ @param A, B: Two strings. @return: The length of longest common subsequence of A and B. """ def longestCommonSubsequence(self, A, B): if not A or not B: return 0 lenA, lenB = len(A), len(B) lcs = [[0 for i in xrange(1 + lenA)] for j in xrange(1 + lenB)] for i in xrange(1, 1 + lenA): for j in xrange(1, 1 + lenB): if A[i - 1] == B[j - 1]: lcs[i][j] = 1 + lcs[i - 1][j - 1] else: lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]) return lcs[lenA][lenB]~~~### C++~~~class Solution {public: /** * @param A, B: Two strings. * @return: The length of longest common subsequence of A and B. */ int longestCommonSubsequence(string A, string B) { if (A.empty()) return 0; if (B.empty()) return 0; int lenA = A.size(); int lenB = B.size(); vector<vector<int> > lcs = \ vector<vector<int> >(1 + lenA, vector<int>(1 + lenB)); for (int i = 1; i < 1 + lenA; i++) { for (int j = 1; j < 1 + lenB; j++) { if (A[i - 1] == B[j - 1]) { lcs[i][j] = 1 + lcs[i - 1][j - 1]; } else { lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]); } } } return lcs[lenA][lenB]; }};~~~### Java~~~ public class Solution { /** * @param A, B: Two strings. * @return: The length of longest common subsequence of A and B. */ public int longestCommonSubsequence(String A, String B) { if (A == null || A.length() == 0) return 0; if (B == null || B.length() == 0) return 0; int lenA = A.length(); int lenB = B.length(); int[][] lcs = new int[1 + lenA][1 + lenB]; for (int i = 1; i < 1 + lenA; i++) { for (int j = 1; j < 1 + lenB; j++) { if (A.charAt(i - 1) == B.charAt(j - 1)) { lcs[i][j] = 1 + lcs[i - 1][j - 1]; } else { lcs[i][j] = Math.max(lcs[i - 1][j], lcs[i][j - 1]); } } } return lcs[lenA][lenB]; }}~~~### 源码分析注意 Python 中的多维数组初始化方式，不可简单使用`[[0] * len(A)] * len(B)]`, 具体原因是因为 Python 中的对象引用方式 [Stackoverflow](#)。### 复杂度分析两重for 循环，时间复杂度为 O(lenA×lenB)O(lenA \times lenB)O(lenA×lenB), 使用了二维数组，空间复杂度也为 O(lenA×lenB)O(lenA \times lenB)O(lenA×lenB). ### Reference- Stackoverflow> . [Python multi-dimensional array initialization without a loop - Stack Overflow](http://stackoverflow.com/questions/3662475/python-multi-dimensional-array-initialization-without-a-loop)[ ↩](# "Jump back to footnote [Stackoverflow] in the text.")
