Distinct Subsequences 118
Question
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).
Example
Given S = "rabbbit", T = "rabbit", return 3.
Challenge
Do it in O(n2) time and O(n) memory.
O(n2) memory is also acceptable if you do not know how to optimize memory.
Solution
ds[i][j]表示T的前j个字符能够在S的前i个字符中能够取的子字符串的数量。
初始化:
ds[i][0]=1,即T的前0个字符总是能够成为S的一个子字符串
ds[0][j]=0(j>0),即不可能有非空字符串是S的前0个字符的子字符串
状态函数:
如果S.charAt(i-1)==T.charAt(j-1):
ds[i][j]=ds[i-1][j]+ds[i-1][j-1]
否则:
ds[i][j]=ds[i-1][j]
即对于S的第i位字符有两种情况:第一种为不能和T的j位字符匹配,则其值为S的前i-1位字符和T的前j位字符的匹配数;第二种情况为可以和T的第j位字符匹配,则其值为S的前i-1位字符和T的前j位字符的匹配数+S的前i-1位字符和T的前j-1位字符的匹配数。
代码如下:
public class Solution {
/**
* @param S, T: Two string.
* @return: Count the number of distinct subsequences
*/
public int numDistinct(String S, String T) {
// write your code here
if(S == null || T == null){
return 0;
}
int n = S.length();
int m = T.length();
int[][] ds = new int[n + 1][m + 1];
for(int i = 0; i <= n; i++){
ds[i][0] = 1;
}
for(int j = 1; j <= m; j++){
ds[0][j] = 0;
}
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
ds[i][j] = ds[i - 1][j];
if(S.charAt(i - 1) == T.charAt(j - 1)){
ds[i][j] += ds[i - 1][j - 1];
}
}
}
return ds[n][m];
}
}