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.
地址：https://oj.leetcode.com/problems/distinct-subsequences/
算法：这道题的题目描述的不是很清楚。按照题目给出的例子，应该是在S中寻找等于T的子序列，然后求这样的子序列的个数。我是用动态规划解决的，用二维dp来存储子问题的解，其中dp[i][j]表示子问题(T[0~i],S[0~j])的解，这样我们就可以按行优先来完成各个子问题，其中如果T[i]==S[j]，那么dp[i][j]=dp[i][j-1] + dp[i-1][j-1]；否则，dp[i][j]=dp[i][j-1]。其中第一行跟第一列都可以实现初始化。代码：
1 class Solution {
2 public:
3 int numDistinct(string S, string T) {
4 if (S.empty() || T.empty()){
5 return 0;
6 }
7 int len_S = S.size();
8 int len_T = T.size();
9 vector<int> temp(len_S);
10 vector<vector<int> > dp(len_T,temp);
11 if(T[0] == S[0]) dp[0][0] = 1;
12 else dp[0][0] = 0;
13 for(int i = 1; i < len_S; ++i){
14 if(T[0] == S[i]){
15 dp[0][i] = dp[0][i-1] + 1;
16 }else{
17 dp[0][i] = dp[0][i-1];
18 }
19 }
20 for(int i = 1; i < len_T; ++i){
21 dp[i][0] = 0;
22 }
23 for(int i = 1; i < len_T; ++i){
24 for(int j = 1; j < len_S; ++j){
25 if(T[i] != S[j]){
26 dp[i][j] = dp[i][j-1];
27 }else{
28 dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
29 }
30 }
31 }
32 return dp[len_T-1][len_S-1];
33 }
34 };
