115. Distinct Subsequences
转自 ： https://blog.csdn.net/feliciafay/article/details/42959119
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: 3Explanation: 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: 5Explanation: 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^^^
DP题目
dp[i][j]表示T的从0开始长度为i的子串和S的从0开始长度为j的子串的匹配的个数。
比如, dp[2][3]表示T中的ra和S中的rab的匹配情况。
(1)显然，至少有dp[i][j] = dp[i][j - 1].
比如, 因为T 中的"ra" 匹配S中的 "ra"， 所以dp[2][2] = 1 。 显然T 中的"ra" 也匹配S中的 "rab"，所以s[2][3] 至少可以等于dp[2][2]。
(2) 如果T[i-1] == S[j-1]， 那么dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)；
比如, T中的"rab"和S中的"rab"显然匹配，
根据(1)， T中的"rab"显然匹配S中的“rabb”，所以dp[3][4] = dp[3][3] = 1，
根据(2), T中的"rab"中的b等于S中的"rab1b2"中的b2, 所以要把T中的"rab"和S中的"rab1"的匹配个数累加到当前的dp[3][4]中。 所以dp[3][4] += dp[2][3] = 2;
(3) 初始情况是
dp[0][0] = 1; // T和S都是空串.
dp[0][1 ... S.length() ] = 1; // T是空串，S只有一种子序列匹配。
dp[1 ... T.length() ][0] = 0; // S是空串，T不是空串，S没有子序列匹配。
class Solution {
public:
int numDistinct(string s, string t) {
int slen = s.length();
int tlen = t.length();
vector> dp(slen + 1, vector(tlen + 1, 0));
dp[0][0] = 1;
for (int i = 1; i <= slen; i++)
dp[i][0] = 1;
for (int i = 1; i <= slen; ++i)
for (int j = 1; j <= tlen; ++j)
{
dp[i][j] = dp[i - 1][j];
if (s[i - 1] == t[j - 1])
{
dp[i][j] += dp[i - 1][j - 1];
}
}
return dp[slen][tlen];
}
};