5.2.1.1 Distinct Subsequences

一、题目

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.

给定两个字符串S和T，求S有多少个不同的子串与T相同。S的子串定义为在S中任意去掉0个或者多个字符形成的串。

二、解题思路

动态规划，设dp[i][j] 是从字符串S[0...i]中删除几个字符得到字符串T[0...j]的不同的删除方法种类，动态规划方程如下

• 如果S[i] = T[j], dp[i][j] = dp[i-1][j-1]+dp[i-1][j]

• 如果S[i] 不等于 T[j], dp[i][j] = dp[i-1][j]

• 初始条件：当T为空字符串时，从任意的S删除几个字符得到T的方法为1

• dp[0][0] = 1; // T和S都是空串.

  dp[1 ... S.length() - 1][0] = 1; // T是空串，S只有一种子序列匹配。

  dp[0][1 ... T.length() - 1] = 0; // S是空串，T不是空串，S没有子序列匹配。

三、解题代码

public int numDistincts(String S, String T)
{
  int[][] table = new int[S.length() + 1][T.length() + 1];

  for (int i = 0; i < S.length(); i++)
      table[i][0] = 1;

  for (int i = 1; i <= S.length(); i++) {
      for (int j = 1; j <= T.length(); j++) {
          if (S.charAt(i - 1) == T.charAt(j - 1)) {
              table[i][j] += table[i - 1][j] + table[i - 1][j - 1];
          } else {
              table[i][j] += table[i - 1][j];
          }
      }
  }

  return table[S.length()][T.length()];
}