LeetCode Easy Implement strStr()

最新推荐文章于 2021-02-27 02:02:17 发布

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本文深入探讨了strStr()函数的两种实现方法，详细解释了如何在主字符串haystack中查找子字符串needle首次出现的位置，包括特殊情况处理，如空字符串和子串不存在的情况。

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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().

Solution1

public class Solution1 {
    /**
     * 先找到第一个匹配的字符，接着对余下子串进行匹配，
     * 失配则在主串中寻找下一个匹配到子串第一个字符的位置
     * 特殊情况：
     * 1. 子串为空或为空串->返回0
     * 2. 主串为空或为空串->返回-1
     * @param haystack
     * @param needle
     * @return 子串在主串中第一次出现的位置
     */
    public int strStr(String haystack, String needle) {
        if (needle == null || needle.equals("")) {
            return 0;
        }
        if (haystack == null || haystack.equals("")) {
            return -1;
        }
        int start = 0;
        while (start <= haystack.length() - needle.length() &&
                needle.charAt(0) != haystack.charAt(start++));
        if (start > 0) start--;
        if (needle.charAt(0) != haystack.charAt(start)) return -1;
        for (int i = 1; i < needle.length(); i++) {
            if (start + i >= haystack.length()) return -1;
            if (needle.charAt(i) != haystack.charAt(start + i)) {
                start++;
                while (start <= haystack.length() - needle.length()) {
                    if (haystack.charAt(start) == needle.charAt(0)) {
                        break;
                    } else {
                        start++;
                    }
                }
                if (start > haystack.length() - needle.length()) return -1;
                i = 0;
            }
        }
        return start;
    }
}

Solution2

public class Solution2 {
    public int strStr(String haystack, String needle) {
        if (needle == null || needle.length() == 0) return 0;
        for (int i = 0; i <= haystack.length() - needle.length(); i++) {
            if (haystack.substring(i, i + needle.length())
            	.equals(needle)) return i;
        }
        return -1;
    }
}