LeetCode:Distinct Subsequences

最新推荐文章于 2020-06-06 05:36:44 发布

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最新推荐文章于 2020-06-06 05:36:44 发布

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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.

题意：给定两个字符串S和T，找出S中删除若干个字符后，和T相等的字符串的个数。

例如：S="rabbbit"，T="rabbit"，则“ra_bbit" "rab_bit" rabb_it，一共三个（其中‘_’表示删除的字符）

解题思路：用经典的dp来解，构建一个二维数组table[i][j]（其中T的长度为行，S的长度为列），每一个元素表示在T[0...i]、S[0...j]中的解决方案的个数，然后按照以下规则进行递推：

1）i=0，则table[i][j]=1。因为i=0它表示T为空，此时无论S为多少，都存在S=""的情况（删除S中所有字符）。

2）i>0，若T[i] != S[j]，则没有新的解决方案，矩阵依旧和以前一样：table[i][j] = table[i][j-1]

若T[i] == S[j]，则有新的解决方案，其一，它包括table[i-1][j-1]方案的个数；其二，它还包括table[i][j-1]方案的个数（因为把S[j]删掉，它也应该具有和前者一样多的解决方案）；

因此，i>0时的解决方案一共有：table[i][j] = table[i-1][j-1] + table[i][j-1];

int numDistinct(string S, string T) 
{
	int m = T.length(), n = S.length();
	if (m > n)
		return 0;

	int** table = new int *[m + 1];
	for (int i = 0; i <= m; i++)
		table[i] = new int[n+1];

	for (int j = 0; j <= n; j++)
		table[0][j] = 1;

	for (int i = 1; i <= m; i++)
		table[i][0] = 0;

	for (int i = 1; i <= m; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			if (T[i-1] != S[j-1])
				table[i][j] = table[i][j - 1];
			else
				table[i][j] = table[i][j - 1] + table[i - 1][j - 1];
		}
	}

	int res = table[m][n];
	for (int i = 0; i <= m; i++)
		delete[] table[i];

	delete[] table;
	
	return res;
}