28. Implement strStr()

Description

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf()).

Tags:** Two Pointers, String

题解

思路1

题意是从主串中找到子串的索引,如果找不到则返回-1,当子串长度大于主串,直接返回-1,然后我们只需要遍历比较即可。

func strStr(haystack string, needle string) int {
    //    检查数据
    if len(haystack) < len(needle) {
        return -1
    }
    if len(needle) == 0 {
        return 0
    }

    for i := 0; i <= len(haystack)-len(needle); i++ {
        j := 0
        for ; j < len(needle); j++ {
            if haystack[i+j] != needle[j] {
                break
            }
        }

        if j == len(needle) {
            return i
        }
    }
    return -1
}

思路 2

KMP 算法

func strStr(haystack string, needle string) int {
    prefixTable := calcPrefixTable(needle)

    i, j := 0, 0
    for i < len(haystack) && j < len(needle) {
        if -1 == j || haystack[i] == needle[j] {
            i++
            j++
        } else {
            if j == 0 {
                i++
            } else {
                j = prefixTable[j]
            }
        }
    }
    if j == len(needle) {
        return i - j
    }
    return -1
}

func calcPrefixTable(str string) []int {
    next := make([]int, len(str)+1)
    length := 0

    for i := 2; i <= len(str); i++ {
        for length > 0 && str[length] != str[i-1] {
            length = next[length]
        }
        if str[length] == str[i-1] {
            length++
        }
        next[i] = length;
    }

    return next
}

结语

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