• KMP String Matching Algorithm


      This is a program based on Knuth-Morris-Pratt String Matching Algorithm (a.k.a KMP algorithm), which can be tapped to solve the problem POJ 3461.

     1 import java.io.*;
     2 import java.util.*;
     3 
     4 class Input {
     5     private Scanner in;
     6     private StringTokenizer tok;
     7     
     8     public Input() {
     9         in = new Scanner(new BufferedInputStream(System.in));
    10     }
    11     public String nextString() {
    12         while  (tok==null||!tok.hasMoreTokens()) {
    13             tok = new StringTokenizer(in.nextLine());
    14         } 
    15         return tok.nextToken();
    16     }
    17     public int nextInt() {
    18         while  (tok==null||!tok.hasMoreTokens()) {
    19             tok = new StringTokenizer(in.nextLine());
    20         } 
    21         return Integer.parseInt(tok.nextToken());
    22     }
    23     public double nextDouble() {
    24         while  (tok==null||!tok.hasMoreTokens()) {
    25             tok = new StringTokenizer(in.nextLine());
    26         } 
    27         return Double.parseDouble(tok.nextToken());
    28     }
    29     public void close() {
    30         in.close();
    31     }
    32 }
    33 
    34 public class Main {
    35     
    36     public static int[] preproc(String pattern)  {
    37         int[] dp = new int[pattern.length()];
    38         Arrays.fill(dp,-1);
    39         for (int i=1;i<pattern.length();i++) {
    40             for (int j=i-1;j>=0;j=dp[j]) {
    41                 if (pattern.charAt(dp[j]+1)==pattern.charAt(i)) {
    42                     dp[i] = dp[j]+1;
    43                     break;
    44                 }
    45             }
    46         }
    47         return dp;
    48     }
    49     public static int match(String pattern,String test) {
    50         int[] next = preproc(pattern);
    51         int cnt = 0;
    52         for (int i=0,j=0;i<test.length();) {
    53             if (test.charAt(i)==pattern.charAt(j)) {
    54                 ++i;
    55                 if (++j==pattern.length()) {
    56                     cnt++;
    57                     j = next[j-1]+1;
    58                 }
    59             } else if (j>0) {
    60                 j = next[j-1]+1;
    61             } else {
    62                 i++;
    63             }
    64         }
    65         return cnt;
    66     }
    67     public static void main(String[] args) {
    68         Input in = new Input();
    69         int time = in.nextInt();
    70         for (int t=0;t<time;t++) {
    71             System.out.println(match(in.nextString(),in.nextString()));
    72         }
    73         in.close();
    74     }
    75 }
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  • 原文地址:https://www.cnblogs.com/DevinZ/p/4462154.html
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