115. Distinct Subsequences

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).

Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

题意：

给定字符串S和T，求字符串S中有多少种不同的subsequences能完全match字符串T

思路：

这是一道高频dp题，需要熟练掌握

区分substring(要求连续)和subsequence(不要求连续)

进一步理解题意，

S ： rabbbit  删掉第一个‘b’ ， 可以跟T完全match

S ： rabbbit  删掉第二个‘b’ ， 可以跟T完全match

S ： rabbbit  删掉第三个‘b’ ， 可以跟T完全match

return 3 （number of distinct subsequences）

用dp[i][j]来记录S的子序列跟T匹配的个数

初始化的时候，除了要处理dp[0][0]，还要条件反射的习惯性思考是否需要预处理第一个row : dp[0][j] 和第一个col：dp[i][0]

    T =    0 "r a b b i t"  
        0  1  0 0 0 0 0 0   
 S =   "r  1  1 0 0 0 0 0 
        a  1  1 1 0 0 0 0 
        b  1  1 1 1 0 0 0 
        b  1  1 1 2 1 0 0 
        b  1  1 1 3 ?
        i  1  1
        t  1  1

显然，对于是否需要预处理第一个row : dp[0][j]，发现当S为空，T为任意字符都不可能跟S匹配。对于dp[0][j]不需要多做处理，只保留defalut值为0即可

对于是否需要预处理第一个col：dp[i][0], 发现当T为空，S的当前字符都可以partition into two subsequences : " " + 当前字符，所以dp[i][0] = 1

对于dp[i][j]，

若 s.charAt(i-1) ! = t.charAt(j-1) 则S当前的字符必须删掉，再看S和T是否匹配： dp[i][j] = dp[i-1][j]

若 s.charAt(i-1) == t.charAt(j-1) 则S当前的字符要么删掉：dp[i][j] = dp[i-1][j] ；要么保留：dp[i][j] = dp[i-1][j-1]

代码

 1 class Solution {
 2     public int numDistinct(String s, String t) {
 3         int[][] dp = new int[s.length() +1 ][t.length() + 1];
 4         dp[0][0] = 1;
 5         for(int i = 1; i <= s.length() ; i++){
 6             dp[i][0] = 1;
 7         }    
 8         for(int i = 1; i <= s.length() ; i++){
 9             for(int j = 1; j<=t.length(); j++){
10                 if( s.charAt(i-1) != t.charAt(j-1) ){
11                     dp[i][j] =  dp[i-1][j];
12                 }else{
13                     dp[i][j] =  dp[i-1][j-1] + dp[i-1][j]; 
14                 }
15             }
16         }
17         return dp[s.length()][t.length()];
18     }
19 }

115. Distinct Subsequences

猜你喜欢