[leetcode]115. Distinct Subsequences(Java)

最新推荐文章于 2022-03-13 16:21:53 发布

JacobGo

最新推荐文章于 2022-03-13 16:21:53 发布

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分类专栏： leetcode 文章标签： Distinct Subsequence leetcode

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本文详细解析了LeetCode上的第115题“不同子序列”的解决方案，介绍了如何通过动态规划的方法来计算字符串S中等于目标字符串T的不同子序列的数量。文章提供了两种实现方式，展示了行列交换不影响结果的特点。

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https://leetcode.com/problems/distinct-subsequences/#/description

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.

package go.jacob.day713;

/**
 * 115. Distinct Subsequences
 * @author Jacob
 *
 */
public class Demo2 {
	/*
	 * 行列交换结果是一样的
	 */
	public int numDistinct(String s, String t) {
		if(s==null||t==null)
			return 0;
		int[][] res=new int[s.length()+1][t.length()+1];
		for(int i=0;i<=s.length();i++){
			res[i][0]=1;
		}
		
		for(int i=0;i<s.length();i++){
			for(int j=0;j<t.length();j++){
				if(s.charAt(i)==t.charAt(j))
					res[i+1][j+1]=res[i][j]+res[i][j+1];
				else
					res[i+1][j+1]=res[i][j+1];
			}
		}
		return res[s.length()][t.length()];
	}
	
	public int numDistinct_1(String s, String t) {
		if(s==null||t==null)
			return 0;
		int[][] res=new int[t.length()+1][s.length()+1];
		for(int i=0;i<=s.length();i++){
			res[0][i]=1;
		}
		for(int i=0;i<t.length();i++){
			for(int j=0;j<s.length();j++){
				if(t.charAt(i)==s.charAt(j))
					res[i+1][j+1]=res[i][j]+res[i+1][j];
				else
					res[i+1][j+1]=res[i+1][j];
				
			}
		}
		
		return res[t.length()][s.length()];
	}
}