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
.
设计动态规划法算法,应该循序渐进,先设计二维表或三维表,然后再简化,进一步优化,再简化,不要一步就写最简化的程序。
熟练之后,可以直接设计表,根据表特征,翻译为程序。
熟练递归回溯法有助于动态规划法的深入理解。
//最好理解的二维动态规划法,一次性ac。 int numDistinct2(string S, string T) { vector<vector<int> > ta(T.size()+1, vector<int>(S.size()+1)); for (int i = 0; i <= S.size(); i++) { ta[0][i] = 1; } for (int i = 1; i <= T.size(); i++) { for (int j = 1; j <= S.size(); j++) { if (T[i-1] == S[j-1]) ta[i][j] = ta[i][j-1] + ta[i-1][j-1]; else ta[i][j] = ta[i][j-1]; } } return ta[T.size()][S.size()]; }
//转置行和列的1维表动态规划法-优化 int numDistinct(string S, string T) { vector<int> ta(T.size()+1); int a = 0, b = 1; for (int i = 1; i <= S.size(); i++) { b = 1; for (int j = 1; j <= T.size(); j++) { a = ta[j]; if (S[i-1] == T[j-1]) ta[j] += b; b = a; if (b == 0) break; } } return ta[T.size()]; }
//神奇的通过的程序--逆向填表法 int numDistinct2(string S, string T) { vector<int> ta(T.size()+1); ta[0] = 1; for (int i = 0; i < S.size(); i++) { for (int j = T.size()-1; j >= 0; j--) { ta[j+1] += (S[i]==T[j])*ta[j]; } } return ta[T.size()]; }
//2014-2-17 update int numDistinct(string S, string T) { int *table = new int[T.length()+1]; table[0] = 1; for (int i = 1; i <= T.length(); i++) { table[i] = 0; } for (int i = 0; i < S.length(); i++) { int j = i<T.length()?i:T.length(); for (; j >= 0; j--) { if (T[j] == S[i]) table[j+1] += table[j];//把握全局,逆向填表 } } return table[T.length()]; }
Leetcode Distinct Subsequences 动态规划法活用总结
原文:http://blog.csdn.net/kenden23/article/details/19332545