第42题 Distinct Subsequences_formed from the characters for-优快云博客

本文介绍了一种使用动态规划算法来计算字符串S中字符串T的不同子序列的数量的方法。通过实例演示了如何通过比较两个字符串的字符并跟踪可能的状态转移来解决这个问题。

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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.

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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];
    }
}