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[LeetCode] Distinct Subsequences（DP）
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.
以下是LeetCode Discuss中非常好的解决方法， using O(mn) space and running in O(mn) time
class Solution {public: int numDistinct(string S, string T) { int m = T.length(); int n = S.length(); if (m > n) return 0; // impossible for subsequence vector<vector<int>> path(m+1, vector<int>(n+1, 0)); for (int k = 0; k <= n; k++) path[0][k] = 1; // initialization for (int j = 1; j <= n; j++) { for (int i = 1; i <= m; i++) { path[i][j] = path[i][j-1] + (T[i-1] == S[j-1] ? path[i-1][j-1] : 0); } } return path[m][n];}};
其中 ：each cell in matrix Path[i][j] means the number of distinct subsequences of T.substr(1...i) in S(1...j)。
/**
* Further optimization could be made that we can use only 1D array instead of a
* matrix, since we only need data from last time step.
*/
Updated solution now takes O(n) space
int numDistinct(string S, string T) { int m = T.length(); int n = S.length(); if (m > n) return 0; // impossible for subsequence vector<int> path(m+1, 0); path[0] = 1; // initial condition for (int j = 1; j <= n; j++) { // traversing backwards so we are using path[i-1] from last time step for (int i = m; i >= 1; i--) { path[i] = path[i] + (T[i-1] == S[j-1] ? path[i-1] : 0); } } return path[m];}