标签：equals return not interview java dex imp http index
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().
用substring()来判断
注意：Java 7中，string.substring()的时间复杂度是O(n)，n为子串长度
time: O(mn), space: O(1) -- m: needle length, n: haystack length
class Solution { public int strStr(String haystack, String needle) { if(haystack == null || needle == null) return -1; if(haystack.length() < needle.length()) return -1; if(needle.length() == 0) return 0; int m = needle.length(); for(int i = 0; i <= haystack.length() - m; i++) { if(haystack.substring(i, i + m).equals(needle)) return i; } return -1; } }
标签：equals return not interview java dex imp http index
原文地址：https://www.cnblogs.com/fatttcat/p/10133131.html
