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
.
一開始一直不理解題意.....
動態規劃,定義dp[i][j]爲字符串i變換到j的變換方法。
如果S[i]==T[j],那麼dp[i][j] = dp[i-1][j-1] + dp[i-1][j]。意思是:如果當前S[i]==T[j],那麼當前這個字母即可以保留也可以拋棄,所以變換方法等於保留這個字母的變換方法加上不用這個字母的變換方法。
如果S[i]!=T[i],那麼dp[i][j] = dp[i-1][j],意思是如果當前字符不等,那麼就只能拋棄當前這個字符。
遞歸公式中用到的dp[0][0] = 1,dp[i][0] = 0(把任意一個字符串變換爲一個空串只有一個方法)
class Solution {
public:
int numDistinct(string S, string T) {
if( S.size() == 0 || S.size() < T.size())
return 0;
if( T.size() == 0)
return 1;
vector< vector< int> > dp( S.size()+1, vector< int>( T.size()+1, 0));
dp[0][0] = 1;
for( int i = 0; i <= S.size(); ++i)
dp[i][0] = 1;
for( int i = 1; i <= S.size(); ++i){
for( int j = 1; j <= T.size(); ++j){
if( S[i-1] == T[j-1])
dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
else
dp[i][j] = dp[i-1][j];
}
}
return dp[S.size()][T.size()];
}
};