[leetcode] Distinct Subsequences

最新推荐文章于 2017-07-04 10:43:23 发布

原创最新推荐文章于 2017-07-04 10:43:23 发布 · 726 阅读

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探讨了如何计算一个字符串中特定子串的独特子序列数量，详细解释了使用动态规划解决此类问题的方法。

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.

下面这个思路无法通过大数据：

class Solution {
public:
    int numDistinct(string S, string T) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int lengths=S.size();
		int lengtht=T.size();
		int count=0,before=-1,n=0;
		vector<vector<int> >ret(lengtht,vector<int>());
		for(int i=0 ; i<lengtht ; i++){
			for(int j=0 ; j<lengths ; j++){
				if(T[i]==S[j])
					ret[i].push_back(j);
			}
		}//put all the positions into ret
		n=ret.size();
		DFS(ret,0,n,before,count);
		return count;
	}
	void DFS(vector<vector<int> >ret, int depth, int n, int& before,int& count){
		if(depth==n){
			count++;
			return;
		}
		for(int i=0 ; i<ret[depth].size();i++){
			if(ret[depth][i]>before){
				int temp=before;
				before=ret[depth][i];
				DFS(ret,depth+1,n,before,count);
				before=temp;
			}
		}
	}
};

考虑把DFS的过程变成动态规划？

t[i][j]=the number of distinct subsequences of string T of length i in string S of length j. So finally the program returns t[t.length][s.length]

If T[i]!=S[j], then t[i][j]=the number of distinct subsequences of string T of length i in string S of length j-1=t[i][j-1] (the left item in the matrix)

If T[i]==S[j], then t[i][j]=the number of distinct subsequences of string T of length i in string S of length j-1 + the number of distinct subsequences of string T of length i-1 in string S of length j-1 = t[i][j-1] + t[i-1][j-1] (the left item in the matrix+the left up item)

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> t(T.length()+1,vector<int>(S.length()+1));
        
        for(int i=0;i<=T.length();i++) t[i][0]=0;
        for(int i=0;i<=S.length();i++) t[0][i]=0;
        
        for(int i=1;i<=S.length();i++) {
            if(T[0]==S[i-1])
                t[1][i]=t[1][i-1]+1;
            else
                t[1][i]=t[1][i-1];    
        }
        
        
        for(int i=2;i<=T.length();i++) {
            for(int j=1; j<=S.length();j++) {
                if(T[i-1]==S[j-1])
                    t[i][j]=t[i-1][j-1]+t[i][j-1];
                else
                    t[i][j]=t[i][j-1];
            }
        }
        
        return t[T.length()][S.length()];
    }
};