115 Distinct Subsequences 不同的子序列
Description:
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).
It's guaranteed the answer fits on a 32-bit signed integer.
Example:
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
^^^
题目描述:
给定一个字符串 S 和一个字符串 T，计算在 S 的子序列中 T 出现的个数。
一个字符串的一个子序列是指，通过删除一些（也可以不删除）字符且不干扰剩余字符相对位置所组成的新字符串。（例如，"ACE" 是 "ABCDE" 的一个子序列，而 "AEC" 不是）
题目数据保证答案符合 32 位带符号整数范围。
示例 :
示例 1：
输入：S = "rabbbit", T = "rabbit"
输出：3
解释：
如下图所示, 有 3 种可以从 S 中得到 "rabbit" 的方案。
(上箭头符号 ^ 表示选取的字母)
rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
示例 2：
输入：S = "babgbag", T = "bag"
输出：5
解释：
如下图所示, 有 5 种可以从 S 中得到 "bag" 的方案。
(上箭头符号 ^ 表示选取的字母)
babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^
思路:
动态规划
dp[i][j]表示 s[:i]中包含 t[:j]的子序列个数
转移方程: 当 s[i] == t[j], dp[i + 1][j + 1] = dp[i][j] + dp[i][j + 1]
否则 dp[i + 1][j + 1] = dp[i + 1][j]
时间复杂度O(mn), 空间复杂度O(max(m, n))
代码:
C++:
class Solution
{
public:
int numDistinct(string s, string t)
{
vector dp(t.size() + 1, 0);
dp[0] = 1;
for (int i = 0; i < s.size(); i++) for (int j = t.size() - 1; j > -1; j--) if (s[i] == t[j]) dp[j + 1] += dp[j];
return (int)dp.back();
}
};
Java:
class Solution {
public int numDistinct(String s, String t) {
int dp[][] = new int[s.length() + 1][t.length() + 1], m = s.length(), n = t.length();
for (int i = 0; i < m + 1; i++) dp[i][0] = 1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (s.charAt(i) == t.charAt(j)) dp[i + 1][j + 1] = dp[i][j] + dp[i][j + 1];
else dp[i + 1][j + 1] = dp[i][j + 1];
}
}
return dp[m][n];
}
}
Python:
class Solution:
def numDistinct(self, s: str, t: str) -> int:
dp = [[1] + [0] * len(t) for _ in range(len(s) + 1)]
for i in range(len(s)):
for j in range(len(t)):
dp[i + 1][j + 1] = dp[i][j] + dp[i][j + 1] if s[i] == t[j] else dp[i][j + 1]
return dp[-1][-1]