JavaScript/Search/StringSearch.js at master · asyncv8/JavaScript · GitHub

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* String Search

function makeTable (str) {

// create a table of size equal to the length of `str`

// table[i] will store the prefix of the longest prefix of the substring str[0..i]

const table = new Array(str.length)

let maxPrefix = 0

// the longest prefix of the substring str[0] has length

table[0] = 0

// for the substrings the following substrings, we have two cases

for (let i = 1; i < str.length; i++) {

// case 1. the current character doesn't match the last character of the longest prefix

while (maxPrefix > 0 && str.charAt(i) !== str.charAt(maxPrefix)) {

// if that is the case, we have to backtrack, and try find a character that will be equal to the current character

// if we reach 0, then we couldn't find a character

maxPrefix = table[maxPrefix - 1]

}

// case 2. The last character of the longest prefix matches the current character in `str`

if (str.charAt(maxPrefix) === str.charAt(i)) {

// if that is the case, we know that the longest prefix at position i has one more character.

// for example consider `.` be any character not contained in the set [a.c]

// str = abc....abc

// consider `i` to be the last character `c` in `str`

// maxPrefix = will be 2 (the first `c` in `str`)

// maxPrefix now will be 3

maxPrefix++

// so the max prefix for table[9] is 3

}

table[i] = maxPrefix

}

return table

}

// Find all the words that matches in a given string `str`

export function stringSearch (str, word) {

// find the prefix table in O(n)

const prefixes = makeTable(word)

const matches = []

// `j` is the index in `P`

let j = 0

// `i` is the index in `S`

let i = 0

while (i < str.length) {

// Case 1. S[i] == P[j] so we move to the next index in `S` and `P`

if (str.charAt(i) === word.charAt(j)) {

i++

j++

}

// Case 2. `j` is equal to the length of `P`

// that means that we reached the end of `P` and thus we found a match

// Next we have to update `j` because we want to save some time

// instead of updating to j = 0 , we can jump to the last character of the longest prefix well known so far.

// j-1 means the last character of `P` because j is actually `P.length`

// e.g.

// S = a b a b d e

// P = `a b`a b

// we will jump to `a b` and we will compare d and a in the next iteration

// a b a b `d` e

// a b `a` b

if (j === word.length) {

matches.push(i - j)

j = prefixes[j - 1]

// Case 3.

// S[i] != P[j] There's a mismatch!

} else if (str.charAt(i) !== word.charAt(j)) {

// if we found at least a character in common, do the same thing as in case 2

if (j !== 0) {

j = prefixes[j - 1]

} else {

// else j = 0, and we can move to the next character S[i+1]

i++

}

return matches

}

// stringSearch('Hello search the position of me', 'pos')

JavaScript/Search/StringSearch.js at master · asyncv8/JavaScript · GitHub

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