115. Distinct Subsequences

最新推荐文章于 2021-08-15 20:07:33 发布

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本文介绍了一种使用动态规划算法解决字符串子序列计数问题的方法。通过构建二维DP数组，跟踪源字符串S中目标字符串T的所有可能子序列，解决了如何高效计算不同子序列数量的问题。

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问题描述
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.
解决思路
动态规划。dp[i][j] 表示s[0…i-1]中子串t[0…j-1]出现的次数。
初始化状态: d[i][0] = 1
状态转移方程：
若s[i-1] == t[j-1]:
dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
否则：
dp[i][j] = dp[i-1][j]
代码

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.length();
        if (m == 0)
            return 0;
        int n = t.length();
        if (n == 0)
            return 1;
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        for (int i = 0; i <= m; ++i)
            dp[i][0] = 1;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++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[m][n];
    }
};