LeetCode 115. Distinct Subsequences

题目

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.

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思路

这道题目描述的不是很清楚，但从例子来看，题目意思是，S序列中只删除某些元素能变换到T序列，有几种变换方法。利用动规解决这道题目。状态res[i][j]表示到S的第i个元素和T的第j个元素时有多少种变换方式。

当S[i] == T[j]时，S[i]可能有两种操作：当S[i-1]==T[j]时，S[i]可以被删除，令S[i-1]匹配T[j]；当S[i-1]不等于T[j]时，S[i]被保留，用S[i-1]匹配T[j-1]，S[i]匹配T[j]， 所以有res[i][j]=res[i-1][j] +res[i-1][j-1]。

当S[i]不等于T[j]时，删除S[i]，有res[i][j]=res[i-1][j]

基于以上两种情况的讨论，可以得到状态转移方程：res[i][j] = ( S[i]==T[j] ? res[i-1][j-1] : 0) +res[i-1][j]。

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代码

class Solution {public:    int **res;    int numDistinct(string s, string t) {        if(s.size() < t.size())            return 0;        int** res=new int*[s.size()+1];        for(int k=0;k<=s.size();k++)         {             res[k]= new int[t.size()+1];         }        for(int i = 0; i<=s.size(); i++)            res[i][0] = 1;        for(int i = 1; i<=t.size(); i++)            res[0][i] = 0;        for(int j = 1; j <= t.size(); j++)        {            for(int i = 1; i <= s.size();i++)            {                if(s[i-1] == t[j-1])                    res[i][j] = res[i-1][j-1] + res[i-1][j];                else                     res[i][j] = res[i-1][j];            }        }        return res[s.size()][t.size()];    }};