dp/distinct-subsequences

Distinct Subsequences

描述

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.

分析

设状态为f(i,j)，表示T[0,j]在S[0,i]里出现的次数。首先，无论S[i]和T[j]是否相等，若不使用S[i]，则f(i,j)=f(i-1,j)；若S[i]==T[j]，则可以使用S[i]，此时f(i,j)=f(i-1,j)+f(i-1, j-1)。

代码

// Distinct Subsequences
// 二维动规+滚动数组
// 时间复杂度O(m*n)，空间复杂度O(n)
public class Solution {
    public int numDistinct(String s, String t) {
        int[] f = new int[t.length() + 1];
        f[0] = 1;
        for (int i = 0; i < s.length(); ++i) {
            for (int j = t.length() - 1; j >= 0; --j) {
                f[j + 1] += s.charAt(i) == t.charAt(j) ? f[j] : 0;
            }
        }

        return f[t.length()];
    }
}

// Distinct Subsequences
// 二维动规+滚动数组
// 时间复杂度O(m*n)，空间复杂度O(n)
class Solution {
public:
    int numDistinct(const string &S, const string &T) {
        vector<int> f(T.size() + 1);
        f[0] = 1;
        for (int i = 0; i < S.size(); ++i) {
            for (int j = T.size() - 1; j >= 0; --j) {
                f[j + 1] += S[i] == T[j] ? f[j] : 0;
            }
        }

        return f[T.size()];
    }
};