Distinct Subsequences
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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"
Return 3
.
这个题目可以用动态规划来解,既然是动态规划,所以关键是要找到递推关系。
我们设S(i)表示第i个字符之前的S的子串(不包含第i个字符),同理设T(j)表示第j个字符之前的T的子串。
设C(i,j)表示S(i)中有T(j)的个数。
于是就有递推关系:
if (当第S(i)的最后一个字符与T(j)的最后一个字符相等时) C(i,j)=C(i-1,j)+C(i-1,j-1);
else C(i,j)=C(i-1,j)
上面对于if语句成立时,S(i)字符串可以分为俩类
一是S(i)包含结果中一定含有最后一个字符的情况:共有C(i-1,j-1)
二是S(i)包含C(j)中不包含最后一个字符的情况:C(i-1,j)
在编写程序时自然可以利用一个二维数组table[i][j]来表示C(i,j)的值。
java code:
public class DistinctSubsequences {public static int numDistincts(String S, String T) {int[][] table = new int[S.length() + 1][T.length() + 1];for (int i = 0; i <= S.length(); i++) {table[i][0] = 1;// 初始化S到T的空子串为1}for (int i = 1; i <= S.length(); i++) {for (int j = 1; j <= T.length(); j++) {if (S.charAt(i - 1) == T.charAt(j - 1)) {table[i][j] = table[i - 1][j - 1] + table[i - 1][j];} else {table[i][j] = table[i - 1][j];}}}return table[S.length()][T.length()];}public static void main(String[] args) {// String S = "b", T = "b";// String S = "abc", T = "";String S = "rabbbit", T = "rabbit";System.out.println(numDistincts(S, T));}}
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