LeetCode Distinct Subsequences(动态规划)

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.

题意：给出两个字符串s,t，求t在s中子串的个数（注意是非连续的）

思路：用dp(i,j)表示字符串t[1..j]在s[1..i]中出现的子串的个数，其状态转移方程为

如果s[i]==t[j]，有dp(i,j)=dp(i-1,j-1)+dp(i-1,j)

如果s[i]不等于t[j],有dp(i,j)=dp(i-1,j)

注意初始值为dp[0..len(s)][0] = 1,而dp[0][1..len(t)] = 0

代码如下：

 

public class Solution{    public int numDistinct(String s, String t)    {        int len1 = s.length(), len2 = t.length();        int[][] dp = new int[len1 + 1][len2 + 1];        dp[0][0] = 1;        for (int i = 0; i < len2; i++)        {            dp[0][i + 1] = 0;        }        for (int i = 0; i < len1; i++)        {            dp[i + 1][0] = 1;        }        for (int i = 0; i < len1; i++)        {            for (int j = 0; j < len2; j++)            {                if (s.charAt(i) == t.charAt(j))                {                    dp[i + 1][j + 1] = dp[i][j] + dp[i][j + 1];                }                else                {                    dp[i + 1][j + 1] = dp[i][j + 1];                }            }        }        return dp[len1][len2];    }}

动态规划的滚动数组实现如下：

public class Solution{    public int numDistinct(String s, String t)    {        int len1 = s.length(), len2 = t.length();        int[][] dp = new int[2][len2 + 1];        dp[0][0] = 1;        for (int i = 0; i <= len1; i++)        {            for (int j = 0; j <= len2; j++)            {               if (i == 0 && j > 0) {                   dp[i & 1][j] = 0;               } else if (i > 0 && j == 0) {                   dp[i & 1][j] = 1;               } else if (i > 0 && j > 0) {                   if (s.charAt(i - 1) == t.charAt(j - 1)) {                       dp[i & 1][j] = dp[(i - 1) & 1][j] + dp[(i - 1) & 1][j - 1];                   } else {                       dp[i & 1][j] = dp[(i - 1) & 1][j];                   }               }            }        }        return dp[len1 & 1][len2];    }}