115. Distinct Subsequences

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115. Distinct Subsequences

题目

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

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. 

解析

• 此题花费很多时间，对递推公式理解不清楚，用一维表示减少空间
• 对比最大公共子序列和子串

//115. Distinct Subsequences
class Solution_115 {
public:

	int numDistinct(string S, string T) { //母串和子串匹配的次数

		int lenx = T.size(); //子串
		int leny = S.size(); //母串
		if (lenx==0||leny==0)
		{
			return 0;
		}

		vector<vector<int> > dp(leny + 1, vector<int>(lenx + 1, 0));
		for (int i = 0; i <= leny;i++) //遍历母串
		{
			for (int j = 0; j <= lenx;j++) //遍历子串
			{
				if (j==0)
				{
					dp[i][j] = 1; //当子串长度为0时，所有次数都是1
					continue;
				}
				if (i>=1&&j>=1)
				{
					if (S[i - 1] == T[j - 1]) //当前母串和子串当前元素相等
					{
						dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
					}
					else
					{
						dp[i][j] = dp[i-1][j];
					}
				}
				
			}
		}
		return dp[leny][lenx];
	}

	int numDistinct2(string s, string t) {
		vector<int> match(t.size() + 1);
		match[0] = 1;
		for (int i = 0; i<s.size(); i++)
		for (int j = t.size(); j>0; j--)
			match[j] += (t[j - 1] == s[i] ? match[j - 1] : 0);
		return match[t.size()];
	}


	int numDistinct1(string S, string T) {  //bug.计算最长公共子序列
	//测试用例:
	//	"ddd", "dd"
	//	对应输出应该为:3
	//  你的输出为 : 2
		int lenx = S.size();
		int leny = T.size();
		if (lenx==0||leny==0)
		{
			return 0;
		}

		vector<vector<int>> vecs(leny+1, vector<int>(lenx+1, 0));
		for (int i = 1; i <= T.size();i++) //行
		{
			for (int j = 1; j <=lenx;j++) //列
			{
				if (S[j-1]==T[i-1])
				{
					vecs[i][j] = vecs[i-1][j - 1] + 1;
				}
				else
				{
					vecs[i][j] = max(vecs[i - 1][j], vecs[i][j - 1]);
				}
			}
		}

		int cnt = 0;
		if (vecs[leny][lenx] > 0){
			cnt++;
			for (int i = lenx - 1; i > 0; i--)
			{
				if (vecs[leny][i] == vecs[leny][lenx])
				{
					cnt++;
				}
			}
		}

		return cnt;
	}
};

题目来源

• 115. Distinct Subsequences
• 牛客网讨论