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28. Find the Index of the First Occurrence in a String

Description

Given two strings needle and haystack, return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

 

Example 1:

Input: haystack = "sadbutsad", needle = "sad"
Output: 0
Explanation: "sad" occurs at index 0 and 6.
The first occurrence is at index 0, so we return 0.

Example 2:

Input: haystack = "leetcode", needle = "leeto"
Output: -1
Explanation: "leeto" did not occur in "leetcode", so we return -1.

 

Constraints:

  • 1 <= haystack.length, needle.length <= 104
  • haystack and needle consist of only lowercase English characters.

Solutions

Solution 1: Traversal

We compare the string needle with each character of the string haystack as the starting point. If we find a matching index, we return it directly.

Assuming the length of the string haystack is $n$ and the length of the string needle is $m$, the time complexity is $O((n-m) \times m)$, and the space complexity is $O(1)$.

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class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        n, m = len(haystack), len(needle)
        for i in range(n - m + 1):
            if haystack[i : i + m] == needle:
                return i
        return -1

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class Solution {
    public int strStr(String haystack, String needle) {
        if ("".equals(needle)) {
            return 0;
        }

        int len1 = haystack.length();
        int len2 = needle.length();
        int p = 0;
        int q = 0;
        while (p < len1) {
            if (haystack.charAt(p) == needle.charAt(q)) {
                if (len2 == 1) {
                    return p;
                }
                ++p;
                ++q;
            } else {
                p -= q - 1;
                q = 0;
            }

            if (q == len2) {
                return p - q;
            }
        }
        return -1;
    }
}

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class Solution {
private:
    vector<int> Next(string str) {
        vector<int> n(str.length());
        n[0] = -1;
        int i = 0, pre = -1;
        int len = str.length();
        while (i < len) {
            while (pre >= 0 && str[i] != str[pre])
                pre = n[pre];
            ++i, ++pre;
            if (i >= len)
                break;
            if (str[i] == str[pre])
                n[i] = n[pre];
            else
                n[i] = pre;
        }
        return n;
    }

public:
    int strStr(string haystack, string needle) {
        if (0 == needle.length())
            return 0;

        vector<int> n(Next(needle));

        int len = haystack.length() - needle.length() + 1;
        for (int i = 0; i < len; ++i) {
            int j = 0, k = i;
            while (j < needle.length() && k < haystack.length()) {
                if (haystack[k] != needle[j]) {
                    if (n[j] >= 0) {
                        j = n[j];
                        continue;
                    } else
                        break;
                }
                ++k, ++j;
            }
            if (j >= needle.length())
                return k - j;
        }

        return -1;
    }
};

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func strStr(haystack string, needle string) int {
    n, m := len(haystack), len(needle)
    for i := 0; i <= n-m; i++ {
        if haystack[i:i+m] == needle {
            return i
        }
    }
    return -1
}

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function strStr(haystack: string, needle: string): number {
    const m = haystack.length;
    const n = needle.length;
    for (let i = 0; i <= m - n; i++) {
        let isEqual = true;
        for (let j = 0; j < n; j++) {
            if (haystack[i + j] !== needle[j]) {
                isEqual = false;
                break;
            }
        }
        if (isEqual) {
            return i;
        }
    }
    return -1;
}

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impl Solution {
    pub fn str_str(haystack: String, needle: String) -> i32 {
        let haystack = haystack.as_bytes();
        let needle = needle.as_bytes();
        let m = haystack.len();
        let n = needle.len();
        let mut next = vec![0; n];
        let mut j = 0;
        for i in 1..n {
            while j > 0 && needle[i] != needle[j] {
                j = next[j - 1];
            }
            if needle[i] == needle[j] {
                j += 1;
            }
            next[i] = j;
        }
        j = 0;
        for i in 0..m {
            while j > 0 && haystack[i] != needle[j] {
                j = next[j - 1];
            }
            if haystack[i] == needle[j] {
                j += 1;
            }
            if j == n {
                return (i - n + 1) as i32;
            }
        }
        -1
    }
}

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/**
 * @param {string} haystack
 * @param {string} needle
 * @return {number}
 */
var strStr = function (haystack, needle) {
    const slen = haystack.length;
    const plen = needle.length;
    if (slen == plen) {
        return haystack == needle ? 0 : -1;
    }
    for (let i = 0; i <= slen - plen; i++) {
        let j;
        for (j = 0; j < plen; j++) {
            if (haystack[i + j] != needle[j]) {
                break;
            }
        }
        if (j == plen) return i;
    }
    return -1;
};

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public class Solution {
    public int StrStr(string haystack, string needle) {
        for (var i = 0; i < haystack.Length - needle.Length + 1; ++i)
        {
            var j = 0;
            for (; j < needle.Length; ++j)
            {
                if (haystack[i + j] != needle[j]) break;
            }
            if (j == needle.Length) return i;
        }
        return -1;
    }
}

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class Solution {
    /**
     * @param String $haystack
     * @param String $needle
     * @return Integer
     */
    function strStr($haystack, $needle) {
        $strNew = str_replace($needle, '+', $haystack);
        $cnt = substr_count($strNew, '+');
        if ($cnt > 0) {
            for ($i = 0; $i < strlen($strNew); $i++) {
                if ($strNew[$i] == '+') {
                    return $i;
                }
            }
        } else {
            return -1;
        }
    }
}

Solution 2: Rabin-Karp String Matching Algorithm

The Rabin-Karp algorithm essentially uses a sliding window combined with a hash function to compare the hashes of fixed-length strings, which can reduce the time complexity of comparing whether two strings are the same to $O(1)$.

Assuming the length of the string haystack is $n$ and the length of the string needle is $m$, the time complexity is $O(n+m)$, and the space complexity is $O(1)$.

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func strStr(haystack string, needle string) int {
    n, m := len(haystack), len(needle)
    sha, target, left, right, mod := 0, 0, 0, 0, 1<<31-1
    multi := 1
    for i := 0; i < m; i++ {
        target = (target*256%mod + int(needle[i])) % mod
    }
    for i := 1; i < m; i++ {
        multi = multi * 256 % mod
    }

    for ; right < n; right++ {
        sha = (sha*256%mod + int(haystack[right])) % mod
        if right-left+1 < m {
            continue
        }
        // 此时 left~right 的长度已经为 needle 的长度 m 了，只需要比对 sha 值与 target 是否一致即可
        // 为避免 hash 冲突，还需要确保 haystack[left:right+1] 与 needle 相同
        if sha == target && haystack[left:right+1] == needle {
            return left
        }
        // 未匹配成功，left 右移一位
        sha = (sha - (int(haystack[left])*multi)%mod + mod) % mod
        left++
    }
    return -1
}

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function strStr(haystack: string, needle: string): number {
    const m = haystack.length;
    const n = needle.length;
    const next = new Array(n).fill(0);
    let j = 0;
    for (let i = 1; i < n; i++) {
        while (j > 0 && needle[i] !== needle[j]) {
            j = next[j - 1];
        }
        if (needle[i] === needle[j]) {
            j++;
        }
        next[i] = j;
    }
    j = 0;
    for (let i = 0; i < m; i++) {
        while (j > 0 && haystack[i] !== needle[j]) {
            j = next[j - 1];
        }
        if (haystack[i] === needle[j]) {
            j++;
        }
        if (j === n) {
            return i - n + 1;
        }
    }
    return -1;
}

Solution 3: KMP String Matching Algorithm

Assuming the length of the string haystack is $n$ and the length of the string needle is $m$, the time complexity is $O(n+m)$, and the space complexity is $O(m)$.

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