[LeetCode]Distinct Subsequences,解题报告
Posted 低调小一
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[LeetCode]Distinct Subsequences,解题报告相关的知识,希望对你有一定的参考价值。
题目
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.
思路1
开始很容易想到深搜,通过flags数组做标记位,得到子串的个数代码:
import java.util.Scanner;
public class DistinctSubsequences
private static int disNum = 0;
public static int numDistinct(String S, String T)
int[] flags = new int[S.length()];
int num = 0;
dfs(num, flags, 0, 0, S, T);
return disNum;
public static void dfs(int num, int[] flags, int indexS, int indexT, String S, String T)
if (num == T.length())
disNum++;
else
for (int i = indexS; i < S.length(); i ++)
if (S.charAt(i) == T.charAt(indexT) && flags[i] == 0)
flags[i] = 1;
num++;
dfs(num, flags, i + 1, indexT + 1, S, T);
flags[i] = 0;
num--;
public static void main(String[] args)
Scanner cin = new Scanner(System.in);
while (cin.hasNext())
String S = cin.nextLine();
String T = cin.nextLine();
disNum = 0;
int res = numDistinct(S, T);
System.out.println(res);
cin.close();
但是在大集合的时候 Time Limit Exceeded
思路2
既然简单的深搜超时,只能考虑稍微复杂一点的DP了。可以参考动态规划经典的例子,最长公共子序列。这里我采用二维数组int[][] dp来记录匹配子序列的个数,则状态方程为:
dp[0][0] = 1, T和S均为空串 dp[0][1..S.length() - 1] = 1, T为空串,S只有一种子序列匹配 dp[1..T.length() - 1][0] = 0, S为空串 dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)
代码:
public class Solution
public static int numDistinct(String S, String T)
if (S == null || S.length() == 0)
return 0;
int[][] dp = new int[T.length() + 1][S.length() + 1];
dp[0][0] = 1;
for (int i = 1; i <= S.length(); i++)
dp[0][i] = 1;
for (int i = 1; i <= T.length(); i++)
dp[i][0] = 0;
for (int i = 1; i <= T.length(); i++)
for (int j = 1; j <= S.length(); j++)
if (T.charAt(i - 1) == S.charAt(j - 1))
dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1];
else
dp[i][j] = dp[i][j - 1];
return dp[T.length()][S.length()];
以上是关于[LeetCode]Distinct Subsequences,解题报告的主要内容,如果未能解决你的问题,请参考以下文章
[LeetCode]题解(python):115-Distinct Subsequences
[动态规划] leetcode 115 Distinct Subsequences
LeetCode Distinct Subsequences
[LeetCode]Distinct Subsequences,解题报告