[Leetcode] Distinct Subsequences

最新推荐文章于 2017-07-04 10:43:23 发布

liuwenyirachel

最新推荐文章于 2017-07-04 10:43:23 发布

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分类专栏： Leetcode 文章标签： dp

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本文介绍了一种计算给定字符串中另一个字符串的唯一子序列数量的方法，通过使用动态规划来解决这一问题。

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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.

比较简单的二维动规，假设distinct[i][j]表示S的前i位和T的前j位的 Distinct Subsequences 个数。

状态方程：

distinct[i][j] = (S[i]==T[j])?distinct[i-1][j]+distinct[i-1][j-1]:distinct[i-1][j]

要初始化一下边界。

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(S==""||T=="")
        {
            return 0;
        }
        
        vector<vector<int> > distinct (S.size(),vector<int>(T.size()));
        
        for(int i=0;i<S.size();i++)
        {
            if(T[0]==S[i])
            {
                distinct[i][0] = (i==0)?1:distinct[i-1][0]+1;
            }
            else
            {
                distinct[i][0] = (i==0)?0:distinct[i-1][0];
            }
        }
        
        for(int i=1;i<T.size();i++)
        {
            distinct[0][i] = 0;
        }
        
        for(int i=1;i<S.size();i++)
        {
            for(int j=1;j<T.size();j++)
            {
                if(j>i)//剪枝
                {
                    distinct[i][j] = 0;
                    continue;
                }
                
                if(S[i]==T[j])
                {
                    distinct[i][j] = distinct[i-1][j] + distinct[i-1][j-1];
                }
                else
                {
                    distinct[i][j] = distinct[i-1][j];
                }
            }
        }
        
        return distinct[S.size()-1][T.size()-1];
    }
};