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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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题意：求S到T的的可能变换。
思路：动规
设：S = "rabbbit"，T="rabbit"
dp[T.length()+1][S.length()+1]；
手动计算可得：
j 0 1 2 3 4 5 6 7
i S r a b b b i t
0 T 1 1 1 1 1 1 1 1
1 r 0 1 1 1 1 1 1 1
2 a 0 0 1 1 1 1 1 1
3 b 0 0 0 1 2 3 3 3
4 b 0 0 0 0 1 3 3 3
5 i 0 0 0 0 0 0 3 3
6 t 0 0 0 0 0 0 0 3
结果即为：3（==dp[T.length()][S.length()] ）；
观察上面dp表中的划横线部分数字，生成过程可以得到：
if(T[i] == S[j]) dp[i][j] = dp[i-1][j-1] + dp[i][j-1];
else dp[i][j] = dp[i][j-1];
即：
当T[i] == S[j]，当前字符可以保留也可以舍弃；
当T[i] != S[j]时，当前字符只能舍弃。
i==0时，表示T为空，这时只有一种变换可能，即去掉S中全部字符。
java code:
public class Solution {
public int numDistinct(String s, String t) {
int m = t.length();
int n = s.length();
int[][] dp = new int[m+1][n+1];
for(int i=0;i<=m;i++) dp[i][0] = 0;
for(int j=0;j<=n;j++) dp[0][j] = 1;
for(int i=1;i<=m;i++) {
for(int j=1;j<=n;j++) {
if(s.charAt(j-1)==t.charAt(i-1))
dp[i][j] = dp[i-1][j-1] + dp[i][j-1];
else
dp[i][j] = dp[i][j-1];
}
}
return dp[m][n];
}
}LeetCode:Distinct Subsequences
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原文地址：http://blog.csdn.net/itismelzp/article/details/51531210
