[LeetCode]Distinct Subsequences,解題報告
阿新 • • 發佈:2017-05-25
blog ati i++ without || args ack 個數 etc
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.
開始非常easy想到深搜,通過flags數組做標記位。得到子串的個數
代碼:
可是在大集合的時候 Time Limit Exceeded
這裏我採用二維數組int[][] dp來記錄匹配子序列的個數,則狀態方程為:
dp[0][0] = 1, T和S均為空串 dp[0][1..S.length() - 1] = 1, T為空串,S僅僅有一種子序列匹配 dp[1..T.length() - 1][0] = 0, S為空串 dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)
代碼:
題目
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
代碼:
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(); } }
可是在大集合的時候 Time Limit Exceeded
思路2
既然簡單的深搜超時,僅僅能考慮略微復雜一點的DP了。能夠參考動態規劃經典的樣例。最長公共子序列。
這裏我採用二維數組int[][] dp來記錄匹配子序列的個數,則狀態方程為:
dp[0][0] = 1, T和S均為空串 dp[0][1..S.length() - 1] = 1, T為空串,S僅僅有一種子序列匹配 dp[1..T.length() - 1][0] = 0, S為空串 dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)
代碼:
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()]; } }
[LeetCode]Distinct Subsequences,解題報告