[leetcode] Distinct Subsequences

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.

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int sLen = S.size();
        int tLen = T.size();
        
        if(sLen==0 ||tLen == 0)
            return 0;
            
        int value[sLen+1][tLen+1];
        memset(value,0,sizeof(int)*(sLen+1)*(tLen+1));
        
        for(int i = 0; i< sLen; i++){
            if (S[i] == T[0])
                value[i+1][1]=value[i][1]+1;
            else
                value[i+1][1]=value[i][1];
        }
        
        for(int i = 0; i< sLen; i++)
            for(int j = 1; j<tLen; j++){
                if (S[i]==T[j])
                    value[i+1][j+1] = value[i][j]+value[i][j+1];
                else
                    value[i+1][j+1] = value[i][j+1];
            }
        return value[sLen][tLen];
    }
};

用DPmethod

Value [i][j] = value[i-1][j-1] + value[i-1][j]   if (s[i]==s[j])
                =  value[i-1][j]                         o.w.

dp式子最開始列錯了。寫成Value [i][j] = value[i][j-1]       o.w., 體會爲什麼錯。（DP式子取決於題目的需求，在本題中，不是求T的substring在S中的情況，而是求整個T的。）

Dp的矩陣到底申請slen還是slen+1取決於第一行/列的賦值情況： 如果第一行/列的取值可以定義用const賦值，就可以申請slen; 如果第一行/列的值需要之前值爲參考，則申請+1大小的空間