[LeetCode] Distinct Subsequences

hale1007

于 2014-10-23 00:25:22 发布

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本文介绍了一种用于计算一个字符串作为另一个字符串的子序列出现次数的算法。通过使用动态规划的方法，该算法能有效地解决这一问题。示例中给出了具体的实现代码，包括初始化条件、状态转移方程等关键步骤。

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Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

class Solution {
public:
    int numDistinct(string S, string T) {
        int m = S.size(), n = T.size();
        vector<vector<int>> res(m+1, vector<int>(n+1, 0));
        for(int i = 0; i < m+1; i++)
            res[i][0] = 1;
        for(int i = 1; i < m+1; i++)
        {
            for(int j = 1; j < n+1; j++)
            {
                if(S[i-1] == T[j-1])
                    res[i][j] = res[i-1][j-1] + res[i-1][j];
                else
                    res[i][j] = res[i-1][j];
            }
        }
        return res[m][n];
    }
};