[leetcode]115. Distinct Subsequences(Java)

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];elseres[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];elseres[i+1][j+1]=res[i+1][j];}}return res[t.length()][s.length()];}}