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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).
Example 1:
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
^^^ ^^^
Example 2:
Input: S = "babgbag", T = "bag"
Output: 5
Explanation:
As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)
babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^
子序列是不用连续的(字串要连续)
这种题一般都需要转换思考方式,用 S 中的字符,按顺序匹配 T 中的字符。
表示 S前i个字符 与 T前j个字符 匹配的个数
- ,则 不匹配 ,
- ,则 既可以不匹配 ,也可以匹配 ,即
边界情况是T串不取时,匹配数都为1。
class Solution {
public:
int numDistinct(string s, string t) {
int n = s.size(), m = t.size();
vector<vector<int>>f(n + 1, vector<int>(m + 1, 0));
for (int i = 0; i <= n; i ++ ) f[i][0] = 1;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
{
f[i][j] = f[i-1][j];
if (s[i-1] == t[j-1]) f[i][j] += f[i-1][j-1];
}
return f[n][m];
}
};