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LeetCode 115 Distinct Subsequences

2024-08-02 02:28:02   221人阅读

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.

思路1：回溯法，结果TLE
public class Solution {
	public static int counter = 0;

	public int numDistinct(String S, String T) {
		counter = 0;
		Distinct(S, T);
		return counter;
	}

	public void Distinct(String S, String T) {
		if (T.length() > 0) {
			Pattern p = Pattern.compile(T.substring(0, 1));
			Matcher m = p.matcher(S);
			while (m.find()) {
				Distinct(S.substring(m.start() + 1), T.substring(1));
			}
		} else {
			counter++;
		}
	}
}
思路2：动态规划，用二维数组result[T.length() + 1][S.length() + 1]来存储T.subString(0,i)在S.subString(0,j)的数目，

如果T.charAt(i - 1) == S.charAt(j - 1)则 result[i][j] = result[i - 1][j - 1] + result[i][j - 1];

否则result[i][j] = result[i][j - 1];

不过要先算一下T.charAt(0) == S.charAt(i-1)若成立，则result[1][i] = result[1][i - 1] + 1;否则result[1][i] = result[1][i - 1];

public class Solution {
	public int numDistinct(String S, String T) {
		int[][] result = new int[T.length() + 1][S.length() + 1];
		for (int i = 1; i <=S.length(); i++) {
			if (T.charAt(0) == S.charAt(i-1)) {
				result[1][i] = result[1][i - 1] + 1;
			} else {
				result[1][i] = result[1][i - 1];
			}
		}
		for (int i = 2; i <= T.length(); i++) {
			for (int j = 1; j <= S.length(); j++) {
				if (T.charAt(i - 1) == S.charAt(j - 1))
					result[i][j] = result[i - 1][j - 1] + result[i][j - 1];
				else
					result[i][j] = result[i][j - 1];
			}
		}

		return result[T.length()][S.length()];
	}
}


LeetCode 115 Distinct Subsequences


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