LeetCode 28：实现strStr()（Implement strStr()）解法汇总

最新推荐文章于 2025-03-25 21:58:09 发布

Ubuntu1996

最新推荐文章于 2025-03-25 21:58:09 发布

阅读量164

点赞数

分类专栏： LeetCode刷题题解记录文章标签： LeetCode 实现strStr() Implement strStr()

本文链接：https://blog.youkuaiyun.com/Utopiabuntu/article/details/92840523

版权

LeetCode刷题题解记录专栏收录该内容

54 篇文章

订阅专栏

博客围绕LeetCode的实现strStr()函数展开，该函数用于返回needle在haystack中首次出现的索引，若不存在则返回 -1。给出了示例，还探讨了needle为空字符串时返回 0 的情况，最后作者分享了自己顺序遍历比较的简单做法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

文章目录

My Solution

更多LeetCode题解

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().

My Solution

我的做法很简单，顺序遍历haystack并与needle比较

class Solution {
public:
	int strStr(string haystack, string needle) {
		int lNeedle = needle.length();
		int lHaystack = haystack.length();
		if (lNeedle == 0) { return 0; }
		for (int i = 0; i <= lHaystack - lNeedle + 1; i++) {
			if (needle.compare(haystack.substr(i, lNeedle)) == 0) {
				return i;
			}
		}
		return -1;
	}
};