LeetCode:Distinct Subsequences

本文轉載自查看原文 2013-11-24 14:50 2416 leetcode/ OJ/ LeetCode

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.

題目大意：刪除S中某些位置的字符可以得到T，總共有幾種不同的刪除方法

設S的長度為lens，T的長度為lent

算法1：遞歸解法，首先，從個字符串S的尾部開始掃描，找到第一個和T最后一個字符相同的位置k，那么有下面兩種匹配：a. T的最后一個字符和S[k]匹配，b. T的最后一個字符不和S[k]匹配。a相當於子問題:從S[0...lens-2]中刪除幾個字符得到T[0...lent-2]，b相當於子問題：從S[0...lens-2]中刪除幾個字符得到T[0...lent-1]。那么總的刪除方法等於a、b兩種情況的刪除方法的和。遞歸解法代碼如下，但是通過大數據會超時：

 1 class Solution {
 2 public:
 3     int numDistinct(string S, string T) {
 4         // IMPORTANT: Please reset any member data you declared, as
 5         // the same Solution instance will be reused for each test case.
 6         return numDistanceRecur(S, S.length()-1, T, T.length()-1);
 7     }
 8     int numDistanceRecur(string &S, int send, string &T, int tend)
 9     {
10         if(tend < 0)return 1;
11         else if(send < 0)return 0;
12         while(send >= 0 && S[send] != T[tend])send--;
13         if(send < 0)return 0;
14         return numDistanceRecur(S,send-1,T,tend-1) + numDistanceRecur(S,send-1,T,tend);
15     }
16 };

View Code

算法2：動態規划，設dp[i][j]是從字符串S[0...i]中刪除幾個字符得到字符串T[0...j]的不同的刪除方法種類，有上面遞歸的分析可知，動態規划方程如下

如果S[i] = T[j], dp[i][j] = dp[i-1][j-1]+dp[i-1][j]
如果S[i] 不等於 T[j], dp[i][j] = dp[i-1][j]
初始條件：當T為空字符串時，從任意的S刪除幾個字符得到T的方法為1

代碼如下：本文地址

 1 class Solution {
 2 public:
 3     int numDistinct(string S, string T) {
 4         // IMPORTANT: Please reset any member data you declared, as
 5         // the same Solution instance will be reused for each test case.
 6         int lens = S.length(), lent = T.length();
 7         if(lent == 0)return 1;
 8         else if(lens == 0)return 0;
 9         int dp[lens+1][lent+1];
10         memset(dp, 0 , sizeof(dp));
11         for(int i = 0; i <= lens; i++)dp[i][0] = 1;
12         for(int i = 1; i <= lens; i++)
13         {
14             for(int j = 1; j <= lent; j++)
15             {
16                 if(S[i-1] == T[j-1])
17                     dp[i][j] = dp[i-1][j-1]+dp[i-1][j];
18                 else dp[i][j] = dp[i-1][j];
19             }
20         }
21         return dp[lens][lent];
22     }
23 };

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