最后一遍-L28-Implement strStr()
Implement strStr().
Return the index of the first occurrence of
needle in haystack, or -1 if needle is not part of haystack.
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().
Example 1:
Input: haystack = "hello", needle = "ll"
Output: 2
Example 2:
Input: haystack = "aaaaa", needle = "bba"
Output: -1
Example 3:
Input: haystack = "", needle = ""
Output: 0
Constraints:
0 <= haystack.length, needle.length <= 5 * 104
haystack and needle consist of only lower-case English characters.
按照题目的意思，进行编码
import util.ListNode;
public class Solution {
public static void main(String[] args) {
System.out.println(strStr("abcdfgderfh", "der"));
}
/**
* we will return 0 when needle is an empty string.
* This is consistent to C's strstr() and Java's indexOf().
*
* @param source target
* 0 <= haystack.length, needle.length <= 5 * 104
* haystack and needle consist of only lower-case English characters.
*
* code: follow the problem
*
* example source="abcfgde" target="de"
* */
public static int strStr(String source, String target) {
if (target == null || target.isEmpty()) {
return 0;
} else {
char targetFirst = target.charAt(0); //'d'
int max = source.length() - target.length();// 5
for (int i = 0; i <= max; i++) {
/** Look for fisrt charscter. */
if (source.charAt(i) != targetFirst) {
while (++i <= max && source.charAt(i) != targetFirst) ;
}
/* Found targetFirst character,now look at the rest of v2*/
if (i <= max) {
int j = i + 1, end = j + target.length() - 1;
for (int k = 1; j < end && source.charAt(j) == target.charAt(k); j++, k++) ;
if (j == end) {
/* Found whole string*/
return i;
}
}
}
}
return -1;
}
}