Distinct Subsequences

最新推荐文章于 2024-03-21 01:52:03 发布

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最新推荐文章于 2024-03-21 01:52:03 发布

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本文介绍了一种计算字符串S中子序列数量的方法，针对给定字符串T，通过动态规划算法实现。详细解释了子序列的概念，并通过实例演示算法应用。

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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.

public class Solution {
    public int numDistinct(String s, String t) {
        int[][] dp = new int[s.length()+1][t.length()+1];
        dp[0][0] = 1;
        for (int i = 1; i <= s.length(); i++) {
			dp[i][0] = 1;
		}
        for (int i = 1; i <= t.length(); i++) {
			dp[0][i] = 0;
		}
        for (int i = 1; i <= s.length(); i++) {
			for (int j = 1; j <= t.length(); j++) {
				dp[i][j] = dp[i-1][j];
				if (s.charAt(i-1) == t.charAt(j-1)) {
					dp[i][j] += dp[i-1][j-1];
				}
			}
		}
        return dp[s.length()][t.length()];
    }
}