LeetCode——Distinct Subsequences

標籤：leetcode

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.

原題連結：https://oj.leetcode.com/problems/distinct-subsequences/

題目：給定一個字串S和字串T，計算T的唯一子序列在S中的個數。

一個字串的子序列是一個新的字串，其從源字串頭部開始，中間可能刪除一些字元，不改變現有字元的相對順序。(例如,"ACE" 是"ABCDE" 的子序列，但 "AEC"不是)

這是一個例子:

S = "rabbbit", T = "rabbit"

返回 3.

思路：動態規劃，dp[i][j]表示T的前j位是S的前i位的子串的情況數。遞推公式是如果S的第i位等於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