动态规划—distinct-subsequences

题目：

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"

Return3.

 

思路：

1. 初始化一个矩阵number[i][j]用来记录字符串T的前j个字符出现在字符串S的前i个字符的次数，当j=0时，令number[i][j]=1；

2. 当S的第i个字符与T的第j个字符不同时，则说明S的第i个字符对number[i][j]没有影响，即number[i][j]=number[i-1][j]；

3. 当S的第i个字符与T的第j个字符不同时，则说明S的第i个字符对number[i][j]有影响，number[i][j]除了要算上原来的number[i-1][j]，还要算上新的可能性，即number[i-1][j-1].

例子

  0 r a b b i t
0 1 0 0 0 0 0 0
r 1 1 0 0 0 0 0
a 1 1 1 0 0 0 0
b 1 1 1 1 0 0 0
b 1 1 1 2 1 0 0
b 1 1 1 3 3 0 0
i 1 1 1 3 3 3 0
t 1 1 1 3 3 3 3

 

代码：

 1     public static int result(String str1, String str2){
 2         int len1 = str1.length(), len2 = str2.length();
 3         int[][] res = new int[len1+1][len2+1];
 4         for(int i=0;i<len1+1;i++){
 5             res[i][0] = 1;
 6         }
 7         for(int i=1;i<len1+1;i++){
 8             for(int j=1;j<len2+1;j++){
 9                 if(str1.charAt(i-1)!=str2.charAt(j-1))
10                     res[i][j] = res[i-1][j];
11                 else
12                     res[i][j] = res[i-1][j]+res[i-1][j-1];
13             }
14             
15         }
16         return res[len1][len2];
17     }

 

posted on 2017-09-01 10:09  一个不会coding的girl  阅读(221)  评论(0)  编辑  收藏  举报

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