【leetcode】115. Distinct Subsequences【java】
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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
.
遇到这种两个串的问题,很容易想到DP。但是这道题的递推关系不明显。可以先尝试做一个二维的表int[][] dp,用来记录匹配子序列的个数(以S="rabbbit"
,T = "rabbit"
为例):
r a b b b i t
1 1 1 1 1 1 1 1
r 0 1 1 1 1 1 1 1
a 0 0 1 1 1 1 1 1
b 0 0 0 1 2 3 3 3
b 0 0 0 0 1 3 3 3
i 0 0 0 0 0 0 3 3
t 0 0 0 0 0 0 0 3
从这个表可以看出,无论T的字符与S的字符是否匹配,dp[i][j] = dp[i][j - 1].就是说,假设S已经匹配了j - 1个字符,得到匹配个数为dp[i][j - 1].现在无论S[j]是不是和T[i]匹配,匹配的个数至少是dp[i][j - 1]。除此之外,当S[j]和T[i]相等时,我们可以让S[j]和T[i]匹配,然后让S[j - 1]和T[i - 1]去匹配。所以递推关系为:
dp[0][0] = 1; // T和S都是空串.
dp[0][1 ... S.length() - 1] = 1; // T是空串,S只有一种子序列匹配。
dp[1 ... T.length() - 1][0] = 0; // S是空串,T不是空串,S没有子序列匹配。
dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0).1 <= i <= T.length(), 1 <= j <= S.length()
Java代码:
public class Solution { public int numDistinct(String s, String t) { int[][] dp = new int[t.length() + 1][s.length() + 1]; dp[0][0] = 1; for (int i = 1; i <= t.length(); i++) { dp[i][0] = 0; } for (int j = 1; j <= s.length(); j++) { dp[0][j] = 1; } for (int i = 1; i <= t.length(); i++) { for (int j = 1; j <= s.length(); j++) { dp[i][j] = dp[i][j - 1]; if (s.charAt(j - 1) == t.charAt(i - 1)){ dp[i][j] += dp[i - 1][j - 1]; } } } return dp[t.length()][s.length()]; }}
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