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go/src/strings/search.go
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Change-Id: I4dc846227af95a73ee9a3074d0c379ff0fa955df
Reviewed-on: https://go-review.googlesource.com/115798
Reviewed-by: Ian Lance Taylor <iant@golang.org>
Run-TryBot: Ian Lance Taylor <iant@golang.org>
2018-06-01 21:52:00 +00:00

125 lines
4.2 KiB
Go

// Copyright 2012 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
package strings
// stringFinder efficiently finds strings in a source text. It's implemented
// using the Boyer-Moore string search algorithm:
// https://en.wikipedia.org/wiki/Boyer-Moore_string_search_algorithm
// https://www.cs.utexas.edu/~moore/publications/fstrpos.pdf (note: this aged
// document uses 1-based indexing)
type stringFinder struct {
// pattern is the string that we are searching for in the text.
pattern string
// badCharSkip[b] contains the distance between the last byte of pattern
// and the rightmost occurrence of b in pattern. If b is not in pattern,
// badCharSkip[b] is len(pattern).
//
// Whenever a mismatch is found with byte b in the text, we can safely
// shift the matching frame at least badCharSkip[b] until the next time
// the matching char could be in alignment.
badCharSkip [256]int
// goodSuffixSkip[i] defines how far we can shift the matching frame given
// that the suffix pattern[i+1:] matches, but the byte pattern[i] does
// not. There are two cases to consider:
//
// 1. The matched suffix occurs elsewhere in pattern (with a different
// byte preceding it that we might possibly match). In this case, we can
// shift the matching frame to align with the next suffix chunk. For
// example, the pattern "mississi" has the suffix "issi" next occurring
// (in right-to-left order) at index 1, so goodSuffixSkip[3] ==
// shift+len(suffix) == 3+4 == 7.
//
// 2. If the matched suffix does not occur elsewhere in pattern, then the
// matching frame may share part of its prefix with the end of the
// matching suffix. In this case, goodSuffixSkip[i] will contain how far
// to shift the frame to align this portion of the prefix to the
// suffix. For example, in the pattern "abcxxxabc", when the first
// mismatch from the back is found to be in position 3, the matching
// suffix "xxabc" is not found elsewhere in the pattern. However, its
// rightmost "abc" (at position 6) is a prefix of the whole pattern, so
// goodSuffixSkip[3] == shift+len(suffix) == 6+5 == 11.
goodSuffixSkip []int
}
func makeStringFinder(pattern string) *stringFinder {
f := &stringFinder{
pattern: pattern,
goodSuffixSkip: make([]int, len(pattern)),
}
// last is the index of the last character in the pattern.
last := len(pattern) - 1
// Build bad character table.
// Bytes not in the pattern can skip one pattern's length.
for i := range f.badCharSkip {
f.badCharSkip[i] = len(pattern)
}
// The loop condition is < instead of <= so that the last byte does not
// have a zero distance to itself. Finding this byte out of place implies
// that it is not in the last position.
for i := 0; i < last; i++ {
f.badCharSkip[pattern[i]] = last - i
}
// Build good suffix table.
// First pass: set each value to the next index which starts a prefix of
// pattern.
lastPrefix := last
for i := last; i >= 0; i-- {
if HasPrefix(pattern, pattern[i+1:]) {
lastPrefix = i + 1
}
// lastPrefix is the shift, and (last-i) is len(suffix).
f.goodSuffixSkip[i] = lastPrefix + last - i
}
// Second pass: find repeats of pattern's suffix starting from the front.
for i := 0; i < last; i++ {
lenSuffix := longestCommonSuffix(pattern, pattern[1:i+1])
if pattern[i-lenSuffix] != pattern[last-lenSuffix] {
// (last-i) is the shift, and lenSuffix is len(suffix).
f.goodSuffixSkip[last-lenSuffix] = lenSuffix + last - i
}
}
return f
}
func longestCommonSuffix(a, b string) (i int) {
for ; i < len(a) && i < len(b); i++ {
if a[len(a)-1-i] != b[len(b)-1-i] {
break
}
}
return
}
// next returns the index in text of the first occurrence of the pattern. If
// the pattern is not found, it returns -1.
func (f *stringFinder) next(text string) int {
i := len(f.pattern) - 1
for i < len(text) {
// Compare backwards from the end until the first unmatching character.
j := len(f.pattern) - 1
for j >= 0 && text[i] == f.pattern[j] {
i--
j--
}
if j < 0 {
return i + 1 // match
}
i += max(f.badCharSkip[text[i]], f.goodSuffixSkip[j])
}
return -1
}
func max(a, b int) int {
if a > b {
return a
}
return b
}