[LeetCode] 28. Implement strStr() 实现strStr()函数

时间:2024-08-26 00:06:32

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().

这道题让我们在一个字符串中找另一个字符串第一次出现的位置,那首先要做一些判断,如果子字符串为空,则返回0,如果子字符串长度大于母字符串长度,则返回 -1。然后开始遍历母字符串,这里并不需要遍历整个母字符串,而是遍历到剩下的长度和子字符串相等的位置即可,这样可以提高运算效率。然后对于每一个字符,都遍历一遍子字符串,一个一个字符的对应比较,如果对应位置有不等的,则跳出循环,如果一直都没有跳出循环,则说明子字符串出现了,则返回起始位置即可,代码如下:

class Solution {
public:
int strStr(string haystack, string needle) {
if (needle.empty()) return ;
int m = haystack.size(), n = needle.size();
if (m < n) return -;
for (int i = ; i <= m - n; ++i) {
int j = ;
for (j = ; j < n; ++j) {
if (haystack[i + j] != needle[j]) break;
}
if (j == n) return i;
}
return -;
}
};

我们也可以写的更加简洁一些,开头直接套两个 for 循环,不写终止条件,然后判断假如j到达 needle 的末尾了,此时返回i;若此时 i+j 到达 haystack 的长度了,返回 -1;否则若当前对应的字符不匹配,直接跳出当前循环,参见代码如下:

解法二:

class Solution {
public:
int strStr(string haystack, string needle) {
for (int i = ; ; ++i) {
for (int j = ; ; ++j) {
if (j == needle.size()) return i;
if (i + j == haystack.size()) return -;
if (needle[j] != haystack[i + j]) break;
}
}
return -;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/28

类似题目:

Shortest Palindrome

Repeated Substring Pattern

参考资料:

https://leetcode.com/problems/implement-strstr/

https://leetcode.com/problems/implement-strstr/discuss/12807/Elegant-Java-solution

https://leetcode.com/problems/implement-strstr/discuss/12956/C%2B%2B-Brute-Force-and-KMP

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