题目:
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.
class Solution {
public:
int numDistinct(string S, string T) {
int m = S.size();
int n = T.size();
vector<vector<int>> f(m + 1, vector<int>(n + 1, 0));
for (int i = 0; i <= m; i++)
f[i][0] = 1;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; 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[m][n];
}
};
本文介绍了一种用于计算一个字符串作为另一个字符串的子序列出现次数的算法。通过动态规划方法,实现了一个高效的解决方案,该算法能正确处理复杂的字符串匹配问题。
4万+

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