Leetcode 115. Distinct Subsequences

问题描述

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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.

问题分析
字符串匹配，定义状态数组dp[i][j]表示s的前j个字符串可以提取t的前i个字符串的子序列数目。
所以有状态转移方程：
dp[i][j]=dp[i][j-1]+(s[j-1]==t[i-1]?dp[i-1][j-1]:0);
即表示t的第j个字符串可以由s的i-1个字符匹配，或者由s的第i个字符匹配（s[j-1]==t[i-1]）。
初始状态，t为空字符，dp[0][j]=1; s为空字符， dp[i][0]=0;
代码如下：

    public int numDistinct(String S, String T) {            int[][] dp = new int[T.length() + 1][S.length() + 1];            dp[0][0] = 1;            for (int i = 1; i <= T.length(); i++) {//s是空串              dp[i][0] = 0;            }            for (int j = 1; j <= S.length(); j++) {//t是空串              dp[0][j] = 1;            }            for (int i = 1; i <= T.length(); i++) {//t在外层循环                  for (int j = 1; j <= S.length(); j++) {                        dp[i][j] = dp[i][j - 1];                        if (T.charAt(i - 1) == S.charAt(j - 1)) {                            dp[i][j] += dp[i - 1][j - 1];                        }                  }            }            return dp[T.length()][S.length()];          }

参考链接：http://blog.csdn.net/abcbc/article/details/8978146