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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.
题目要求计算T在S的子序列中出现的次数，这里的子序列只要求顺序与在S中的相对顺序一致就行。
思路：使用动态规划解题。
设map[i][j]表示S[0~i]的子序列中出现T[0~j]的次数，则：
1. 如果S[i] != T[j],则map[i][j] = map[i-1][j]
2. 如果S[i] == T[j],则map[i][j] = map[i-1][j-1] + map[i-1][j];
代码如下：
public int numDistinct(String S, String T) { if(S == null || S.length() == 0) return 0; if(T == null || T.length() == 0) return 1; if(S.length() < T.length()) return 0; int map[][] = new int[S.length()][T.length()]; if(S.charAt(0) == T.charAt(0)) map[0][0] = 1; for(int i=1; i<S.length(); i++) { if(S.charAt(i) == T.charAt(0)) map[i][0] = map[i-1][0] + 1; else { map[i][0] = map[i-1][0]; } } for(int i=1; i<S.length(); i++) { for(int j=1; j<=i && j<T.length(); j++) { if(S.charAt(i) == T.charAt(j)) { map[i][j] = map[i-1][j-1] + map[i-1][j]; } else { map[i][j] = map[i-1][j]; } } } return map[S.length()-1][T.length()-1]; }
LeetCode-115 Distinct Subsequences
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原文地址：http://www.cnblogs.com/linxiong/p/4334023.html
