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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.
这题的dp解法要先看基础点的
public class Solution { public int numDistinct(String s, String t) { if(s==null||s.length()<t.length()) return 0; int[][] dp = new int[s.length()+1][t.length()+1]; dp[0][0] = 1; for(int i=1;i<=s.length();i++){ dp[i][0] = 1; // t is empty, matches whatever s is } for(int j=1;j<=t.length();j++){ dp[0][j] = 0; // s is empty, nothing in t can match } for(int i=1;i<=s.length();i++){ for(int j=1; j<=t.length();j++){ //注意这个位置是第i位置，第j位，dp和s，t的位置序号不一致！ // if t is already matched or not within s.subarray（0，i） // consider if t is still within s.subarray（0，i+1）, actually determined by without i // e.g.: s[rato], t[rat] dp[i][j] = dp[i-1][j]; if(s.charAt(i-1)==t.charAt(j-1)){ // in this situation, originally mismatched t[rat‘b‘], due to the addition of ‘b‘ to s[ratk...b], now it is determined by both minus 1 character "b" dp[i][j] += dp[i-1][j-1]; } } } return dp[s.length()][t.length()]; } }
BM 解法
基础
但是这道题我反正理解不了
public class Solution { public int numDistinct(String S, String T) { int[] occurence = new int[T.length() + 1]; occurence[0] = 1; for(int i = 0; i < S.length(); i++){ for(int j = T.length() - 1; j >= 0 ; j--) if(S.charAt(i) == T.charAt(j)){ if(occurence[j] > 0) occurence[j + 1] += occurence[j]; } } return occurence[T.length()]; } }
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原文地址：http://www.cnblogs.com/jiajiaxingxing/p/4565071.html
