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30. Distinct Subsequences
Distinct Subsequences
OJ: https://oj.leetcode.com/problems/distinct-subsequences/
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.
思想：动态规划。 D[i][j] = D[i][j-1] + (T[i-1] == S[j-1] ? D[i-1][j-1] : 0);
// DP: D[i][j] = D[i][j-1] + (T[i-1] == S[j-1] ? D[i-1][j-1] : 0);class Solution {public: int numDistinct(string S, string T) { int m = T.length(); int n = S.length(); if(!m) return 1; if(m > n) return 0; vector<vector<int> > D(m+1, vector<int>(n+1)); for(int i = 1; i <= m; ++i) D[i][0] = 0; for(int i = 0; i <= n; ++i) D[0][i] = 1; for(int i = 0; i < m; ++i) for(int j = 0; j < n; ++j) D[i+1][j+1] = D[i+1][j] + (T[i] == S[j] ? D[i][j] : 0); return D[m][n]; }};改进后：空间复杂度 O(T.size()).
class Solution {public: int numDistinct(string S, string T) { int m = T.length(); vector<int> num(m+1, 0); // num[i] is distinct numbers of T[1,...,i] in string S num[0] = 1; for(int i = 1; i <= S.length(); ++i) for(int j = min(i, m); j >= 1; --j) // key to notice. if(T[j-1] == S[i-1]) num[j] += num[j-1]; // num[j-1] 为上次字符串时，T[1,...,j-1] 的 distinct numbers。 return num[m]; }};
30. Distinct Subsequences
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