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.
分析：此题乍一看看不明确啥意思，后面慢慢理解。就是说T是S的子串，这里的子串是说原字符串删除某些字符后剩余字符的组合，题目是S中通过删除操作获得子串T的数目。如样例中的第一个为删掉第一个b，第二个为删除第二个b，第三个为删除第三个b，因此数量为3.
首先想到的方法是回溯法。可是遇到长串时超时。里面包括太多反复子问题。
代码例如以下：Time Limit Exceeded
class Solution {
public:
int numDistinct(string S, string T) {
num = 0;
if(S.length() < T.length()){
return num;
}
numDistinctCore(S,T,0,0);
return num;
}
void numDistinctCore(string& S,string& T, int si, int tj){
int slen = S.length();
int tlen = T.length();
if(slen-si < tlen -tj){
return;
}
if(tj == tlen){
++num;
}
for(int i = si; i<slen; ++i){
if(S[i] == T[tj]){
numDistinctCore(S,T,i+1, tj+1);
}
}
}
int num;
};考虑到回溯法子问题反复求解。想到利用动态规划的方法。此题有些像求最大公共字串，可是须要改动一下，寻找子问题。
设dp[i][j]为S字符串截止到i时。能将S前面字符串能转换为T截止到j的子串的转换次数。求dp[i][j]时。一种方法转换时即删除S[i]，变回S[i-1][j]的问题。还有一种方法。假设S[i] == T[j]。则转换为dp[i-1][j-1]的问题。
即为同样时为 dp[i][j] = dp[i-1][j] + dp[i-1][j-1] 。不同一时候为 dp[i][j] = dp[i-1][j]
Accepted
class Solution {
public:
int numDistinct(string S, string T) {
int slen = S.length();
int tlen = T.length();
if(slen < tlen){
return 0;
}
int **dp = new int*[slen+1];
for(int i=0; i<slen+1; ++i){
dp[i] = new int[tlen+1];
dp[i][0] = 1;
}
for(int j=1; j<tlen+1; ++j){
dp[0][j] = 0;
}
for(int i=1; i<slen+1; ++i){
for(int j=1; j<tlen+1; ++j){
int temp = dp[i-1][j];
if(S[i-1] == T[j-1]){
temp += dp[i-1][j-1];
}
dp[i][j] = temp;
}
}
int result = dp[slen][tlen];
for(int i=0; i<slen+1; ++i){
delete[] dp[i];
}
delete[] dp;
return result;
}
};