題目:spa
Given a string S and a string T, count the number of distinct subsequences of T in S.code
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).orm
Here is an example:
S = "rabbbit"
, T = "rabbit"
blog
Return 3
. rem
思路:string
動態規劃;關鍵是如何獲得遞推關係,能夠這樣想,設母串的長度爲 j,子串的長度爲 i,咱們要求的就是長度爲 i 的字串在長度爲 j 的母串中出現的次數,設爲 dp[i][j],若母串的最後一個字符與子串的最後一個字符不一樣,則長度爲 i 的子串在長度爲 j 的母串中出現的次數就是母串的前 j - 1 個字符中子串出現的次數,即 dp[i][j] = dp[i][j - 1],若母串的最後一個字符與子串的最後一個字符相同,那麼除了前 j - 1 個字符出現字串的次數外,還要加上子串的前 i - 1 個字符在母串的前 j - 1個字符中出現的次數,即dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]。 it
package dp; public class DistinctSubsequences { public int numDistinct(String s, String t) { int slen = s.length(); int tlen = t.length(); if (tlen > slen) return 0; int[][] dp = new int[slen + 1][tlen + 1]; for (int i = 0; i <= slen; ++i) { for (int j = 0; j <= tlen; ++j) { if (j == 0) { dp[i][j] = 1; continue; } if (j > i) { dp[i][j] = 0; continue; } if (s.charAt(i - 1) == t.charAt(j - 1)) dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1]; else dp[i][j] = dp[i - 1][j]; } } return dp[slen][tlen]; } public static void main(String[] args) { // TODO Auto-generated method stub DistinctSubsequences d = new DistinctSubsequences(); System.out.println(d.numDistinct("rabbbit", "rabbit")); } }