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().
解題大意:
返回neelde在haystack中第一次出現(xiàn)的索引位置痹束，如果不是haystack的一部分吁断，則返回-1念颈。
解題思路：
看完題目就想到了KMP算法泊业，然后開始動(dòng)手實(shí)現(xiàn)。后來發(fā)現(xiàn)python語言有專門的find()函數(shù)用來查找子串咙鞍。
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
return haystack.find(needle)
這樣就解決了苞冯，感覺異常簡單。其實(shí)也可以用python的切片方法來實(shí)現(xiàn)find()的查找功能肯尺，值得注意的是遍歷range(n)的遍歷范圍是0~n-1,這里i的范圍為[0,m-n]閉區(qū)間沃缘。
時(shí)間復(fù)雜度是O(m)
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
m = len(haystack)
n = len(needle)
for i in range(m-n+1):
if haystack[i:i+n] == needle:
return i
return -1
參考：https://blog.csdn.net/fuxuemingzhu/article/details/79254558
