Upgrade dependent version: github.com/open-policy-agent/opa (#5315)
Upgrade dependent version: github.com/open-policy-agent/opa v0.18.0 -> v0.45.0 Signed-off-by: hongzhouzi <hongzhouzi@kubesphere.io> Signed-off-by: hongzhouzi <hongzhouzi@kubesphere.io>
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hongzhouzi
197
vendor/github.com/open-policy-agent/opa/internal/lcss/lcss.go
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197
vendor/github.com/open-policy-agent/opa/internal/lcss/lcss.go
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package lcss
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import "bytes"
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// LongestCommonSubstring returns the longest substring which is present in all the given strings.
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// https://en.wikipedia.org/wiki/Longest_common_substring_problem
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// Not to be confused with the Longest Common Subsequence.
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// Complexity:
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// * time: sum of `n_i*log(n_i)` where `n_i` is the length of each string.
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// * space: sum of `n_i`.
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// Returns a byte slice which is never a nil.
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//
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// ### Algorithm.
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// We build suffix arrays for each of the passed string and then follow the same procedure
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// as in merge sort: pick the least suffix in the lexicographical order. It is possible
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// because the suffix arrays are already sorted.
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// We record the last encountered suffixes from each of the strings and measure the longest
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// common prefix of those at each "merge sort" step.
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// The string comparisons are optimized by maintaining the char-level prefix tree of the "heads"
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// of the suffix array sequences.
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func LongestCommonSubstring(strs ...[]byte) []byte {
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strslen := len(strs)
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if strslen == 0 {
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return []byte{}
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}
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if strslen == 1 {
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return strs[0]
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}
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suffixes := make([][]int, strslen)
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for i, str := range strs {
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suffixes[i] = qsufsort(str)
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}
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return lcss(strs, suffixes)
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}
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func lcss(strs [][]byte, suffixes [][]int) []byte {
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strslen := len(strs)
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if strslen == 0 {
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return []byte{}
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}
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if strslen == 1 {
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return strs[0]
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}
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minstrlen := len(strs[0]) // minimum length of the strings
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for _, str := range strs {
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if minstrlen > len(str) {
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minstrlen = len(str)
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}
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}
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heads := make([]int, strslen) // position in each suffix array
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boilerplate := make([][]byte, strslen) // existing suffixes in the tree
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boiling := 0 // indicates how many distinct suffix arrays are presented in `boilerplate`
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var root charNode // the character tree built on the strings from `boilerplate`
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lcs := []byte{} // our function's return value, `var lcss []byte` does *not* work
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for {
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mini := -1
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var minSuffixStr []byte
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for i, head := range heads {
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if head >= len(suffixes[i]) {
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// this suffix array has been scanned till the end
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continue
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}
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suffix := strs[i][suffixes[i][head]:]
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if minSuffixStr == nil {
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// initialize
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mini = i
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minSuffixStr = suffix
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} else if bytes.Compare(minSuffixStr, suffix) > 0 {
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// the current suffix is the smallest in the lexicographical order
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mini = i
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minSuffixStr = suffix
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}
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}
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if mini == -1 {
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// all heads exhausted
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break
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}
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if boilerplate[mini] != nil {
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// if we already have a suffix from this string, replace it with the new one
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root.Remove(boilerplate[mini])
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} else {
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// we track the number of distinct strings which have been touched
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// when `boiling` becomes strslen we can start measuring the longest common prefix
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boiling++
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}
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boilerplate[mini] = minSuffixStr
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root.Add(minSuffixStr)
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heads[mini]++
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if boiling == strslen && root.LongestCommonPrefixLength() > len(lcs) {
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// all heads > 0, the current common prefix of the suffixes is the longest
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lcs = root.LongestCommonPrefix()
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if len(lcs) == minstrlen {
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// early exit - we will never find a longer substring
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break
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}
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}
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}
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return lcs
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}
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// charNode builds a tree of individual characters.
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// `used` is the counter for collecting garbage: those nodes which have `used`=0 are removed.
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// The root charNode always remains intact apart from `children`.
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// The tree supports 4 operations:
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// 1. Add() a new string.
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// 2. Remove() an existing string which was previously Add()-ed.
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// 3. LongestCommonPrefixLength().
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// 4. LongestCommonPrefix().
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type charNode struct {
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char byte
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children []charNode
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used int
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}
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// Add includes a new string into the tree. We start from the root and
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// increment `used` of all the nodes we visit.
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func (cn *charNode) Add(str []byte) {
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head := cn
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for i, char := range str {
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found := false
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for j, child := range head.children {
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if child.char == char {
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head.children[j].used++
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head = &head.children[j] // -> child
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found = true
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break
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}
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}
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if !found {
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// add the missing nodes one by one
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for _, char = range str[i:] {
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head.children = append(head.children, charNode{char: char, children: nil, used: 1})
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head = &head.children[len(head.children)-1]
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}
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break
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}
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}
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}
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// Remove excludes a node which was previously Add()-ed.
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// We start from the root and decrement `used` of all the nodes we visit.
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// If there is a node with `used`=0, we erase it from the parent's list of children
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// and stop traversing the tree.
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func (cn *charNode) Remove(str []byte) {
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stop := false
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head := cn
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for _, char := range str {
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for j, child := range head.children {
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if child.char != char {
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continue
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}
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head.children[j].used--
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var parent *charNode
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head, parent = &head.children[j], head // shift to the child
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if head.used == 0 {
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parent.children = append(parent.children[:j], parent.children[j+1:]...)
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// we can skip deleting the rest of the nodes - they have been already discarded
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stop = true
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}
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break
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}
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if stop {
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break
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}
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}
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}
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// LongestCommonPrefixLength returns the length of the longest common prefix of the strings
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// which are stored in the tree. We visit the children recursively starting from the root and
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// stop if `used` value decreases or there is more than one child.
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func (cn charNode) LongestCommonPrefixLength() int {
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var result int
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for head := cn; len(head.children) == 1 && head.children[0].used >= head.used; head = head.children[0] {
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result++
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}
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return result
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}
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// LongestCommonPrefix returns the longest common prefix of the strings
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// which are stored in the tree. We compute the length by calling LongestCommonPrefixLength()
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// and then record the characters which we visit along the way from the root to the last node.
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func (cn charNode) LongestCommonPrefix() []byte {
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result := make([]byte, cn.LongestCommonPrefixLength())
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if len(result) == 0 {
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return result
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}
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var i int
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for head := cn.children[0]; ; head = head.children[0] {
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result[i] = head.char
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i++
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if i == len(result) {
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break
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}
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}
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return result
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}
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