【Leetcode】115. Distinct Subsequences

最新推荐文章于 2024-02-14 11:24:21 发布

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最新推荐文章于 2024-02-14 11:24:21 发布

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原文链接：http://www.cnblogs.com/luntai/p/5660500.html

本文探讨了如何解决计数一个字符串在另一个字符串中作为子序列出现次数的问题。通过动态规划的方法，给出了详细的分析过程及代码实现。

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Description:

　　Given two string S and T, you need to count the number of T's subsequences appeared in S. The fucking problem description is so confusing.

Input:

　　String s and t

output:

　　The number

Analysis:

　　It's a dynamic processing problem. I drew the dynamic processing of counting the seq numbers and then got the correct formula by guessing? :) Most times I work out the final formula by deducing! Then i back to think it's real sense in the problem.

dp[i][j] represents the number of subsequences in string T (ending before index i) are appeared in string S (ending before index j). So, dp can be processed by the follow formula:

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

　dp[i][j]

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

BYT:

　　The fucking input size of test cases in serve are ambiguity! So if you create a 2-dimension array in defined size, you will be in trouble. Dynamic structures like vector will be better!

Code:

class Solution {
public:
    int numDistinct(string s, string t) {
        if(s.length() == 0 || t.length() == 0) return 0;
        //int dp[50][10908];
        vector<vector<int>> dp(t.length() + 1, vector<int>(s.length() + 1, 0));
        dp[0][0] = (t[0] == s[0])?1:0;
        for(int i = 1; i < s.length(); i ++){
            if(s[i] == t[0]) dp[0][i] = dp[0][i - 1] + 1;
            else dp[0][i] = dp[0][i - 1];
        }
        for(int i = 1; i < t.length(); i ++){
            dp[i][i - 1] = 0;
            for(int j = i; j < s.length(); j ++){
                dp[i][j] = (t[i] == s[j])? (dp[i][j - 1] + dp[i - 1][j - 1]):dp[i][j - 1];
            }
        }
        return dp[t.length() - 1][s.length() - 1];
    }
};