[LeetCode] 115. Distinct Subsequences

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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.

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找到T在S中有几种子序列表示，如：
S = “rabbbit”, T = “rabbit”

[
rabb b it
rab b bit
ra b bbit
]
三种

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动规解决，找到转移方程：
if (s[i] == t[j]) {
  dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
} else {
  dp[i][j] = dp[i-1][j];
}

dp[i-1][j-1] 表示s[0:i-1]已经有这么多个t[0:j-1]的子序列。
dp[i-1][j] 表示s[0:i-1]已经有这么多个t[0:j]的子序列。
如果s[i] == t[j]， 可以在dp[i-1][j]数量的基础上新增dp[i-1][j-1]个子序列。

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class Solution {public:    int numDistinct(string s, string t) {        int len1 = s.length();        int len2 = t.length();        if (len1 == 0 || len2 == 0) return 0;        vector<vector<int>> dp(len1, vector<int>(len2, 0));        if (s[0] == t[0]) dp[0][0] = 1;        for (int i=1; i<len1; ++i) {            if (s[i] == t[0]) dp[i][0] = dp[i-1][0] + 1;            else dp[i][0] = dp[i-1][0];        }        for (int i=1; i<len1; ++i) {            for (int j=1; j<len2; ++j) {                if (s[i] == t[j]) dp[i][j] = dp[i-1][j-1] + dp[i-1][j];                else dp[i][j] = dp[i-1][j];            }        }        return dp[len1-1][len2-1];    }};