LintCode 77. Longest Common Subsequence 最长公共序列

题目链接

Description
  Given two strings, find the longest common subsequence (LCS).
Your code should return the length of LCS.

Example 1:

Input: "ABCD" and "EDCA"
Output: 1
Explanation:
LCS is 'A' or 'D' or 'C'

Example 2:

Input: "ABCD" and "EACB"
Output: 2
Explanation:
LCS is "AC"

思路：
  这是一个经典的动态规划题，定义dp矩阵，dp[i][j]表示A串匹配到i，B串匹配到j时的最大公共长度。dp有A.length()+1行，有B.length()列。第一行与第一列都为0；
有状态转移方程：

最后返回dp的最大值即为最长公共子串长度，代码如下：

class Solution {
public:
    /**
     * @param A: A string
     * @param B: A string
     * @return: The length of longest common subsequence of A and B
     */
    int longestCommonSubsequence(string &A, string &B) {
        // write your code here
        int m = A.size(), n = B.size();
        if (m == 0 || n==0) return 0;
        //int dp[m+1][n+1];
        std::vector> dp(m+1, vector(n+1, 0));
        for (int i=1; i<=m; ++i) {
            for (int j=1; j<=n; ++j) {
                if (A[i-1] == B[j-1]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                else {
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        return dp[m][n];
    }
};

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• LintCode 79. Longest Common Substring
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