[LeetCode] 28. Implement strStr()

Implement strStr().

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

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().

Example 1:

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

Example 2:

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

Example 3:

Input: haystack = "", needle = ""
Output: 0

Constraints:

• 0 <= haystack.length, needle.length <= 5 * 104
• haystack and needle consist of only lower-case English characters.

实现strStr()函数。

题意是给定一个 haystack 字符串和一个 needle 字符串，在 haystack 字符串中找出 needle 字符串出现的第一个位置 (从0开始)。如果不存在，则返回  -1。这个题有KMP算法的答案但是我这里只给出另一个简单的解法，不算暴力解。

时间O(n)

空间O(1)

JavaScript实现

/**
 * @param {string} haystack
 * @param {string} needle
 * @return {number}
 */
var strStr = function (haystack, needle) {
    // corner case
    if (needle === '') {
        return 0;
    }

    if (needle.length > haystack.length) {
        return -1;
    }

    // normal case
    let m = haystack.length;
    let n = needle.length;
    for (let i = 0; i < m - n + 1; i++) {
        if (haystack.substring(i, i + n) === needle) {
            return i;
        }
    }
    return -1;
};

Java实现

class Solution {
    public int strStr(String haystack, String needle) {
        // corner case
        if (needle == null || needle.length() == 0) {
            return 0;
        }

        if (needle.length() > haystack.length()) {
            return -1;
        }

        // normal case
        int m = haystack.length();
        int n = needle.length();
        for (int i = 0; i < m - n + 1; i++) {
            if (haystack.substring(i, i + n).equals(needle)) {
                return i;
            }
        }
        return -1;
    }
}

LeetCode 题目总结