Distinct Subsequences 不同子序列

最新推荐文章于 2021-03-26 16:50:53 发布

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最新推荐文章于 2021-03-26 16:50:53 发布

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本文链接：https://blog.youkuaiyun.com/nutcc/article/details/10858035

本文介绍了一种高效算法，用于计算一个字符串作为另一个字符串子序列出现的不同方式的数量。通过使用动态规划方法，避免了重复计算，显著提高了算法效率。

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Distinct Subsequences

Oct 19 '12

6266 / 17972

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.

我的很复杂的思路。。。。

package String;

import java.util.HashMap;

/**
 * @Title: Distinct_Subsequences.java
 * @Package String
 * @Description: TODO
 * @author nutc
 * @date 2013-9-1 下午3:16:51
 * @version V1.0
 */
public class Distinct_Subsequences {
	
	public static void main(String argsp[]){
		Distinct_Subsequences l = new Distinct_Subsequences();
		String s = "ddd";
		String t = "";
		System.out.println(l.numDistinct(s, t));
	}

	public int numDistinct(String S, String T) {
		if (S == null && T == null)
			return 0;
		if (S == null)
			return 0;
		if (T == null)
			return 1;
		if (S.equals(T))
			return 1;
		return compare(S, 0, T, 0);

	}

	HashMap<String, Integer> map = new HashMap<String, Integer>();

	public int compare(String s, int i, String t, int j) {
		if (i == s.length())
			return 0;
		if (j == t.length())
			return 1;//
		if (j == t.length() - 1 && i == s.length() - 1) {
			if (s.charAt(i) == t.charAt(j))
				return 1;
			else
				return 0;
		}

		int count = 0;
		if (s.charAt(i) == t.charAt(j)) {
			if (map.get((i + 1) + "-" + (j + 1)) != null)
				count += map.get((i + 1) + "-" + (j + 1));
			else
				count += compare(s, i + 1, t, j + 1);
		}
		if (map.get((i + 1) + "-" + (j)) != null)
			count += map.get((i + 1) + "-" + (j));
		else
			count += compare(s, i + 1, t, j);
		map.put(i + "-" + j, count);
		return count;
	}
}

但是其实想清楚的话是很简单的！

public class Solution {
    public int numDistinct(String S, String T) {
        int m = T.length();
        int n = S.length();
        int[][] dp = new int[m+1][n+1];
        dp[0][0] = 1;
        for(int i=0;i<=m;i++)
            dp[i][0] = 0; //这两个初始值的区别
        for(int j=0;j<=n;j++)
            dp[0][j] = 1; //这两个初始值的区别
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++){
                if (T.charAt(i-1)==S.charAt(j-1))
                    dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
                else
                    dp[i][j] = dp[i][j-1];
            }
        return dp[m][n];
    }
}