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: 3Explanation: As shown below, there are 3 ways you can generate "rabbit" from S. (The caret symbol ^ means the chosen letters)rabbbit^^^^ ^^rabbbit^^ ^^^^rabbbit^^^ ^^^
Example 2:
Input: S ="babgbag", T ="bag" Output: 5Explanation: 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; } };