Problem:
思路：
用DP算法。状态方程为：
s[i-1] == t[j-1] dp[i][j] = dp[i-1][j-1] + dp[i-1][j] //如果相等时，就考虑2种可能情况，包括s[i-1]和不包括s[i-1]的情况。
s[i-1] != t[j-1] dp[i][j] = dp[i-1][j] //不相等时只需考虑不包括s[i-1]的情况。
Solution (C++):
int numDistinct(string s, string t) {
int m = s.size(), n = t.size(); //size()和length()方法均可
vector<vector<long>> dp(m+1, vector<long>(n+1, 0));
for (int i = 0; i < m; ++i) dp[i][0] = 1;
for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
dp[i][j] = dp[i-1][j] + (s[i-1] == t[j-1] ? dp[i-1][j-1] : 0);
return dp[m][n];
}
性能：
Runtime: 12 ms Memory Usage: 15.6 MB
思路：
Solution (C++):
int m = s.size(), n = t.size(); //size()和length()方法均可
vector<long long> dp(n+1, 0);
dp[0]= 1;
for (int i = 1; i <= m; ++i) {
int pre = 1;
for (int j = 1; j <= n; ++j) {
int tmp = dp[j];
dp[j] += (s[i-1] == t[j-1] ? pre : 0);
pre = tmp;
}
}
return dp[n];
性能：
Runtime: 8 ms Memory Usage: 8.9 MB
