LeetCode——Distinct Subsequences

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.

原题链接：https://oj.leetcode.com/problems/distinct-subsequences/

题目：给定一个字符串S和字符串T，计算T的唯一子序列在S中的个数。

一个字符串的子序列是一个新的字符串，其从源字符串头部开始，中间可能删除一些字符，不改变现有字符的相对顺序。(例如,"ACE" 是"ABCDE" 的子序列，但 "AEC"不是)

这是一个例子:

S = "rabbbit", T = "rabbit"

返回 3.

思路：动态规划，dp[i][j]表示T的前j位是S的前i位的子串的情况数。递推公式是如果S的第i位等于T的第j位, dp[i][j] =
dp[i-1][j-1] + dp[i][j-1], 如果不等, dp[i][j] = dp[i][j-1] 。

	public int numDistinct(String S, String T) {
		int lens = S.length();
		int lent = T.length();
		int[][] dp = new int[lent + 1][lens + 1];
		dp[0][0] = 1;
		for (int i = 1; i <= lent; i++)
			dp[i][0] = 0;
		for (int i = 1; i <= lens; i++)
			dp[0][i] = 1;
		for (int i = 1; i <= lent; i++) {
			for (int j = 1; j <= lens; 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] + dp[i][j - 1];
			}
		}
		return dp[lent][lens];
	}

reference ：http://chaoren.is-programmer.com/posts/43585.html

http://blog.csdn.net/abcbc/article/details/8978146

LeetCode——Distinct Subsequences