add longest common subsequence

Author: billryan

dynamic_programming/longest_common_subsequence.md

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+# Longest Common Subsequence
+
+- category: [DP_Two_Sequence]
+
+## 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? Yes
+Example
+For "ABCD" and "EDCA", the LCS is "A" (or "D", "C"), return 1.
+
+For "ABCD" and "EACB", the LCS is "AC", return 2.
+
+Clarification
+What's the definition of Longest Common Subsequence?
+
+https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
+http://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
+
+```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++
+
+```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
+
+```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 $$O(lenA \times lenB)$$, 使用了二维数组，空间复杂度也为 $$O(lenA \times lenB)$$. 
+
+## 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)