Longest Common Subsequence

Problem

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

• Problem link.

Notes

这是个Two Sequence DP问题，定义状态和状态转移方程

1. state: $$f[i][j]$$表示前i个字符配上前j个字符的LCS的长度

2. $$function: f[i][j] = f[i-1][j-1] + 1 // a[i-1] == b[j-1]

    \\= MAX(f[i-1][j], f[i][j-1]) // a[i-1] != b[j-1]$$
   

3. $$intialize: f[i][0] = 0, f[0][j] = 0$$

4. $$answer: f[a.length()][b.length()]$$

Solution

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) {
        // write your code here
        if(A.length() == 0 || B.length()==0) return 0;
        int m = A.length();
        int n = B.length();
        vector<vector<int> > lcs(m+1, vector<int>(n+1,0));
        for(int i=1; i<=m; i++) {
            for(int j=1; j<=n; j++) {
                if(A[i-1] == B[j-1]) {//注意不是比较A[i] and B[j]哦！
                    lcs[i][j] = lcs[i-1][j-1] + 1;
                } else {
                    lcs[i][j] = max(lcs[i][j-1], lcs[i-1][j]);
                }
            }
        }
        return lcs[m][n];
    }
};

需要注意的是lcs状态数组是[m+1][n+1]维的

// Time:  O(m * n)
// Space: O(min(m, n))

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.length() < B.length()) {
            return longestCommonSubsequence(B, A);
        }

        // table[i][j] means the longest length of common subsequence
        // of A[0 : i] and B[0 : j].
        vector<vector<int>> table(2, vector<int>(A.length() + 1, 0));

        // if A[i - 1] != B[j - 1]:
        //     table[i][j] = max(table[i - 1][j], table[i][j - 1])
        // if A[i - 1] == B[j - 1]:
        //     table[i][j] = table[i - 1][j - 1] + 1
        for (int i = 1; i < A.length() + 1; ++i) {
            for (int j = 1; j < B.length() + 1; ++j) {
                if (A[i - 1] != B[j - 1]) {
                    table[i % 2][j] = max(table[(i - 1) % 2][j],
                                       table[i % 2][j - 1]);
                } else {
                    table[i % 2][j] = table[(i - 1) % 2][j - 1] + 1;
                }
            }
        }

        return table[A.length() % 2][B.length()];
    }
};