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.
//Dynamic programming
//dp[i][j] stores the number of distinct subsequences of first j chars in T and first i chars in S
//if(s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
//else dp[i][j] = dp[i - 1][j];
//dp[0][0] = 1
//dp[0][1~m] = 0
//dp[1~n][0] = 1
//reference: http://www.cnblogs.com/ganganloveu/p/3836519.html
class Solution {
public:
int numDistinct(string s, string t) {
int n = s.size();
int m = t.size();
if(n == 0 && m != 0){
return 0;
}
if(n == 0 && m == 0){
return 1;
}
if(m == 0){
return 1;
}
int dp[n + 1][m + 1];
for(int i = 0; i <= n; i++){
dp[i][0] = 1;
}
for(int i = 0; i <= m; i++){
dp[0][i] = 0;
}
dp[0][0] = 1;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
if(s[i - 1] == t[j - 1]){
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
}
else{
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[n][m];
}
};

public class Solution {
public int numDistinct(String s, String t) {
if(s == null || t == null || t.length() == 0) return 0;
int[] dp = new int[t.length()];
for(int i = 0; i<s.length(); i++){
char c = s.charAt(i);
for(int j=dp.length-1; j>=0; j–){
if(c == t.charAt(j)){
dp[j] = dp[j] + (j!=0?dp[j-1]: 1);
}
}
}
return dp[t.length()-1];
}
}
URL: http://traceformula.blogspot.com/2015/08/distinct-subsequences.html
Great! Your solution shrinks the space cost a lot.
I have a solution using less space:
public class Solution {
public int numDistinct(String s, String t) {
if(s == null || t == null || t.length() == 0) return 0;
int[] dp = new int[t.length()];
for(int i = 0; i
=0; j–){if(c == t.charAt(j)){
dp[j] = dp[j] + (j!=0?dp[j-1]: 1);
}
}
}
return dp[t.length()-1];
}
}
URL: http://traceformula.blogspot.com/2015/08/distinct-subsequences.html