LeetCode 115 Distinct Subsequences

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

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Input: S = "babgbag", T = "bag"
Output: 5
Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^

第一到hard难度的题
其实也是一道水题，
首先我用了暴力深搜，果然超时，效率是（最差情况）s的长度：n,t的长度：m

O(n) = n*（n-1）*(n-2)*...(n-m+1)*m

正确的解法应该是前缀和，
遍历t字符串中的每个字符ti 找到ti 在s字符串中的位置si，统计si的前缀和sss,这里的前缀和是值si前面有多少个满足的条件的t的前缀字符串，
O(n)=  n * m

c++

class Solution {
public:
    int result;
    int sss[100005];//ss数组的前缀和数组
    int ss[100005];//当前字符串前面满足条件的前缀字符串的个数
    int numDistinct(string s, string t) {
        int lens = s.length();
        int lent = t.length();
        memset(sss,0,sizeof(sss));
      
        for(int j=0;j<lent;j++)
        {
            memset(ss,0,sizeof(ss));
            for(int i=j;i<lens;i++)
            {
                if(s[i]==t[j])
                {
                    ss[i]=(j==0?1:sss[i-1]);
                }
            }
            for(int i=0;i<lens;i++)
            {
                if(i==0){sss[i]=ss[i];continue;}
                sss[i]=sss[i-1]+ss[i];
            }
        }
        for(int i=0;i<lens;i++)
        {
            result += ss[i];
        }
        return result;
    }
    
};