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70 lines
2.1 KiB
Java
70 lines
2.1 KiB
Java
package com.thealgorithms.searches;
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class KMPSearch {
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int KMPSearch(String pat, String txt) {
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int M = pat.length();
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int N = txt.length();
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// create lps[] that will hold the longest
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// prefix suffix values for pattern
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int lps[] = new int[M];
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int j = 0; // index for pat[]
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// Preprocess the pattern (calculate lps[]
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// array)
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computeLPSArray(pat, M, lps);
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int i = 0; // index for txt[]
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while ((N - i) >= (M - j)) {
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if (pat.charAt(j) == txt.charAt(i)) {
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j++;
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i++;
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}
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if (j == M) {
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System.out.println("Found pattern " + "at index " + (i - j));
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int index = (i - j);
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j = lps[j - 1];
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return index;
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}
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// mismatch after j matches
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else if (i < N && pat.charAt(j) != txt.charAt(i)) {
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// Do not match lps[0..lps[j-1]] characters,
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// they will match anyway
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if (j != 0) j = lps[j - 1]; else i = i + 1;
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}
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}
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System.out.println("No pattern found");
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return -1;
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}
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void computeLPSArray(String pat, int M, int lps[]) {
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// length of the previous longest prefix suffix
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int len = 0;
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int i = 1;
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lps[0] = 0; // lps[0] is always 0
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// the loop calculates lps[i] for i = 1 to M-1
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while (i < M) {
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if (pat.charAt(i) == pat.charAt(len)) {
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len++;
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lps[i] = len;
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i++;
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} else { // (pat[i] != pat[len])
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// This is tricky. Consider the example.
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// AAACAAAA and i = 7. The idea is similar
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// to search step.
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if (len != 0) {
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len = lps[len - 1];
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// Also, note that we do not increment
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// i here
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} else { // if (len == 0)
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lps[i] = len;
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i++;
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}
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}
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}
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}
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}
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// This code has been contributed by Amit Khandelwal.
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