115. Distinct Subsequences

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.

思路及代码：

int numDistinct(string s, string t) {int m = t.size(), n = s.size();if (m > n || m == n && s != t)return 0;vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));dp[0][0] = 1;for (int i = 1; i <= m; i++)dp[i][0] = 0;//这一行可以不写，因为初值就是0;for (int i = 1; i <= n; i++)dp[0][i] = 1;//t为空，不管s多长，匹配的子序列为1；for (int i = 1; i <= m; i++)for (int j = 1; j <= n; j++)//当s[j - 1] ！= t[i - 1]时：既是要求s中前j个与t中前i个匹配的个数。//当s[j - 1] == t[i - 1]时：如果匹配的子序列包含s中的第j个，也一定包含t中第i个，因为他俩都是最后一个。这时情况为dp[i - 1][j - 1]种。//                          如果不包含，那么就在s的钱j-1个字符中去匹配。这时情况为dp[i][j - 1]种。dp[i][j] = dp[i][j - 1] + (s[j - 1] == t[i - 1] ? dp[i - 1][j - 1] : 0);return dp[m][n];}