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().
Tags:** Two Pointers, String
题意是从主串中找到子串的索引，如果找不到则返回-1，当子串长度大于主串，直接返回-1，然后我们只需要遍历比较即可。
class Solution {
public int strStr(String haystack, String needle) {
int l1 = haystack.length(), l2 = needle.length();
if (l1 < l2) return -1;
for (int i = 0; ; i++) {
if (i + l2 > l1) return -1;
for (int j = 0; ; j++) {
if (j == l2) return i;
if (haystack.charAt(i + j) != needle.charAt(j)) break;
}
}
}
}如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：awesome-java-leetcode