116. Distinct Subsequences

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.

题目大意：删除S中某些位置的字符可以得到T，总共有几种不同的删除方法

设S的长度为lens，T的长度为lent

此题总结：DP类的问题，一定要深思熟虑，找出子问题以及子问题和父问题之间的联系。

算法1：递归解法，首先，从个字符串S的尾部开始扫描，找到第一个和T最后一个字符相同的位置k，那么有下面两种匹配：a. T的最后一个字符和S[k]匹配，b. T的最后一个字符不和S[k]匹配。a相当于子问题:从S[0...lens-2]中删除几个字符得到T[0...lent-2]，b相当于子问题：从S[0...lens-2]中删除几个字符得到T[0...lent-1]。那么总的删除方法等于a、b两种情况的删除方法的和。递归解法代码如下，但是通过大数据会超时：

class Solution {public:    int numDistinct(string S, string T) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        return numDistanceRecur(S, S.length()-1, T, T.length()-1);    }    int numDistanceRecur(string &S, int send, string &T, int tend)    {        if(tend < 0)return 1;        else if(send < 0)return 0;        while(send >= 0 && S[send] != T[tend])send--;<pre name="code" class="cpp"><span style="font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 13.9200000762939px;"><span style="white-space:pre"></span>        if(send < 0)return 0;</span>

return numDistanceRecur(S,send-1,T,tend-1) + numDistanceRecur(S,send-1,T,tend); }};

算法2：动态规划，设dp[i][j]是从字符串S[0...i]中删除几个字符得到字符串T[0...j]的不同的删除方法种类，有上面递归的分析可知，动态规划方程如下

• 如果S[i] = T[j], dp[i][j] = dp[i-1][j-1]+dp[i-1][j]
• 如果S[i] 不等于 T[j], dp[i][j] = dp[i-1][j]
• 初始条件：当T为空字符串时，从任意的S删除几个字符得到T的方法为1

代码如下

class Solution {public:    int numDistinct(string S, string T) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        int lens = S.length(), lent = T.length();        if(lent == 0)return 1;        else if(lens == 0)return 0;        int dp[lens+1][lent+1];        memset(dp, 0 , sizeof(dp));        for(int i = 0; i <= lens; i++)dp[i][0] = 1;        for(int i = 1; i <= lens; i++)        {            for(int j = 1; j <= lent; j++)            {                if(S[i-1] == T[j-1])                    dp[i][j] = dp[i-1][j-1]+dp[i-1][j];                else dp[i][j] = dp[i-1][j];            }        }        return dp[lens][lent];    }};