第42题 Distinct Subsequences

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.

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 Dynamic Programming String

Code in Java：

public class Solution {    public int numDistinct(String S, String T) {        //count the number of ways to convert S to T by deleting some characters in S.         if(S.length()==0&&T.length()==0)   return 0;         if(S.length()<T.length()) return 0;         int lenS = S.length(), lenT = T.length();                  //Use opj[i][j] to represent the number of ways to convert S[0...i-1] to T[0...j-1]         //So opj[lenS][lenT] is the total number of ways to convert S to T         int[][] opj = new int[lenS+1][lenT+1];                  //opj[i][0] represents the number of ways to convert S[0...i-1] to an empty string         //so for any string, there is only one ways, which is deleting all the characters         for(int i=0; i<lenS+1; i++)            opj[i][0]=1;                //There are two possibilities when computing opj[i][j] from S[0...i-1] to T[0...j-1]:        //if S[i-1]==T[j-1], we can obtain opj[i][j] by either remaining or deleting S[i-1]        //  if remain S[i-1]:   opj[i][j]=opj[i-1][j-1]         //                      the number of ways from  S[0...i-1] to T[0...j-1]=the number of ways        //                      from S[0...i-2] to T[0...j-2]        //  if delete S[i-1]:   opj[i][j]=opj[i-1][j]        //                      the number of ways from S[0...i-1] to  T[0...j-1]=the number of ways        //                      from S[0...i-2] to T[0...j-1]        //if S[i-1]!=T[j-1], there is only one way to obtain opj[i][j], wo we have        //      opj[i][j] = opj[i-1][j]        //      the number of ways from S[0...i-1] to T[0...j-1]=the nubmer of ways from         //      S[0...i-2] to T[0...j-1]        for(int i=1; i<lenS+1; i++){            for(int j=1; j<lenT+1&&j<=i; j++){                opj[i][j] = S.charAt(i-1)==T.charAt(j-1)?opj[i-1][j-1]:0;                if(i-1>=j)  opj[i][j]+= opj[i-1][j];            }        }        return opj[lenS][lenT];    }}