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
.
题目大意：删除S中某些位置的字符可以得到T，总共有几种不同的删除方法
设S的长度为lens，T的长度为lent
算法1 ：递归解法，首先，从个字符串S的尾部开始扫描，找到第一个和T最后一个字符相同的位置k，那么有下面两种匹配：a. T的最后一个字符和S[k]匹配，b. T的最后一个字符不和S[k]匹配。a相当于子问题:从S[0...lens-2]中删除几个字符得到T[0...lent-2]，b相当于子问题：从S[0...lens-2]中删除几个字符得到T[0...lent-1]。那么总的删除方法等于a、b两种情况的删除方法的和。递归解法代码如下，但是通过大数据会超时：
1
class
Solution {
2
public
:
3
int
numDistinct(
string
S,
string
T) {
4
//
IMPORTANT: Please reset any member data you declared, as
5
//
the same Solution instance will be reused for each test case.
6
return
numDistanceRecur(S, S.length()-
1
, T, T.length()-
1
);
7
}
8
int
numDistanceRecur(
string
&S,
int
send,
string
&T,
int
tend)
9
{
10
if
(tend <
0
)
return
1
;
11
else
if
(send <
0
)
return
0
;
12
while
(send >=
0
&& S[send] != T[tend])send--
;
13
if
(send <
0
)
return
0
;
14
return
numDistanceRecur(S,send-
1
,T,tend-
1
) + numDistanceRecur(S,send-
1
,T,tend);
15
}
16
};
算法2 ：动态规划，设dp[i][j]是从字符串S[0...i]中删除几个字符得到字符串T[0...j]的不同的删除方法种类，有上面递归的分析可知，动态规划方程如下
- 如果S[i] = T[j], dp[i][j] = dp[i-1][j-1]+dp[i-1][j]
- 如果S[i] 不等于 T[j], dp[i][j] = dp[i-1][j]
- 初始条件：当T为空字符串时，从任意的S删除几个字符得到T的方法为1
代码如下： 本文地址
1
class
Solution {
2
public
:
3
int
numDistinct(
string
S,
string
T) {
4
//
IMPORTANT: Please reset any member data you declared, as
5
//
the same Solution instance will be reused for each test case.
6
int
lens = S.length(), lent =
T.length();
7
if
(lent ==
0
)
return
1
;
8
else
if
(lens ==
0
)
return
0
;
9
int
dp[lens+
1
][lent+
1
];
10
memset(dp,
0
,
sizeof
(dp));
11
for
(
int
i =
0
; i <= lens; i++)dp[i][
0
] =
1
;
12
for
(
int
i =
1
; i <= lens; i++
)
13
{
14
for
(
int
j =
1
; j <= lent; j++
)
15
{
16
if
(S[i-
1
] == T[j-
1
])
17
dp[i][j] = dp[i-
1
][j-
1
]+dp[i-
1
][j];
18
else
dp[i][j] = dp[i-
1
][j];
19
}
20
}
21
return
dp[lens][lent];
22
}
23
};
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