[leetcode-115-Distinct Subsequences]

class Solution {
public:
    /*用删除的方法将串s变换到t，计算变换方法数*/
    int numDistinct(string s, string t) {
        if (s.empty() || t.empty())
            return 0;
        else if (s.length() < t.length())
            return 0;
        else
        {
            //动态规划
            int ls = s.length(), lt = t.length();
            /*保存由字符串s(0,i) --> t(0,j)的方法数*/
            vector<vector<int> > dp(ls + 1, vector<int>(lt + 1, 0));
            dp[0][0] = 1;
            for (int i = 0; i < ls; ++i)
            {
                /*s(0,i) 转换为 t(0)的方法数为1*/
                dp[i][0] = 1;
            }//for
            for (int i = 1; i <= ls; ++i)
            {
                for (int j = 1; j <= lt; ++j)
                {
                    /*首先不管当前字符是否相同，为dp[i][j]赋初值*/
                    dp[i][j] = dp[i - 1][j];
                    if (s[i-1] == t[j-1])
                    {
                        /*如果s和t的当前字符相同，有两种选择保留或不保留*/
                        dp[i][j] += dp[i - 1][j - 1];
                    }//if
                }//for
            }//for
            return dp[ls][lt];
        }
    }    
};