LeetCode115. Distinct Subsequences

最新推荐文章于 2024-02-14 11:24:21 发布

原创最新推荐文章于 2024-02-14 11:24:21 发布 · 257 阅读

0 ·

CC 4.0 BY-SA版权

DP 同时被 2 个专栏收录

25 篇文章

订阅专栏

LeetCode每日一题

6 篇文章

订阅专栏

本文介绍了一种使用动态规划解决字符串匹配计数问题的方法。通过两个示例代码详细展示了如何在一维和二维数组中计算源字符串S中有多少不同的子序列等于目标字符串T。一维优化方案减少了空间复杂度。

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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).

Example 1:

Input: S = "rabbbit", T = "rabbit"
Output: 3

思路：

dp[i][j]代表s[0...i]与t[0...j]匹配数

dp[i][j]=dp[i-1][j] //if s[i]!=t[j]

dp[i][j]=dp[i-1][j]+dp[i-1][j-1] //if s[i]==t[j]

AC code

二维写法：

class Solution
{
public:
	int numDistinct(string s, string t)
	{
		if (s.empty() || t.empty()|| s.length() < t.length())
			return 0;

        int ls = s.length(), lt = t.length();
        vector<vector<long long> > dp(ls + 1, vector<long long>(lt + 1, 0));
        dp[0][0] = 1;
        for (int i = 0; i < ls; ++i)
            dp[i][0] = 1;
        for (int i = 1; i <= ls; ++i)
            for (int j = 1; j <= lt; ++j)
            {
                dp[i][j] = dp[i - 1][j];
                if (s[i-1] == t[j-1])
                    dp[i][j] += dp[i - 1][j - 1];
            }
        return dp[ls][lt];
	}
};

一维优化：

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