algorithm - Rabin-Karp字符串匹配的实现(滚动哈希)

Question

我正在尝试实现 Rabin-Karp查找字符串的字符串匹配算法needle在字符串 haystack (返回字符串 haystack 的索引，其中找到了字符串 needle 的匹配项)。我在尝试查找针时遇到错误 c在大海捞针abc .

这是运行查找 c 的代码后输出的样子在abc :

hayHash: 0 and needleHash: 2 at i: 0 remove: a and add: b and new hash is before checking for negatives: 1

hayHash: 1 and needleHash: 2 at i: 1 remove: b and add: c and new hash is before checking for negatives: -50

hayHash: 51 and needleHash: 2

我不明白为什么最后一次哈希重新计数计算为 -50而不是2对于 hayHash 。这里都是hayHash和needleHash应该是仅包含 char 'c' 的计算哈希值，并且两者的值应该相同。但我的代码正在重新计算 hayHash至51 (取消负值之前的-50)而不是2 .

对我的代码中可能存在的问题有什么建议吗？

这是我的代码:

private fun find(haystack: String, needle: String): Int {
    if(needle.length > haystack.length) return -1

    val q = 101
    val d = 256
    var needleHash = 0
    var hayHash = 0
    var hash = 1

    for (i in 0..needle.length)
        hash = (hash * d) % q

    for(i in 0..needle.lastIndex) {
        needleHash = (d * needleHash + (needle[i] - 'a')) % q
        hayHash = (d * hayHash + (haystack[i] - 'a')) % q
    }

    for(i in 0..(haystack.length - needle.length)) {
        println("hayHash: $hayHash and needleHash: $needleHash")
        if(hayHash == needleHash) {
            for(j in 0..needle.lastIndex) {
                if(haystack[i + j] != needle[j])
                    break
                if(j == needle.lastIndex)
                    return i
            }
        }
        if(i == haystack.length - needle.length)
            break
        print("at i: $i remove: ${haystack[i]} and add: ${haystack[i + needle.length]}")   
        hayHash = (d * (hayHash - (haystack[i]  - 'a') * hash) + (haystack[i + needle.length]  - 'a')) % q
        println(" and new hash is before checking for negatives: $hayHash")
        if(hayHash < 0)
            hayHash += q
    }

    return -1
}

Answer 1

初始化 hash 时

for (i in 0..needle.length) 是一个差一错误。您需要 for(i in 0..needle.lastIndex)。