标签:lan count The array targe character inter imp uri
28. Implement strStr()
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().
package leetcode;
public class ImplementStrStr {
@org.junit.Test
public void test() {
String haystack1 = "hello";
String needle1 = "ll";
String haystack2 = "aaaaa";
String needle2 = "bba";
System.out.println(strStr(haystack1, needle1));
System.out.println(strStr(haystack2, needle2));
}
public int strStr(String haystack, String needle) {
char[] source = haystack.toCharArray();
int sourceOffset = 0;
int sourceCount = haystack.length();
char[] target = needle.toCharArray();
int targetOffset = 0;
int targetCount = needle.length();
int fromIndex = 0;
if (fromIndex >= sourceCount) {
return (targetCount == 0 ? sourceCount : -1);
}
if (fromIndex < 0) {
fromIndex = 0;
}
if (targetCount == 0) {
return fromIndex;
}
char first = target[targetOffset];
int max = sourceOffset + (sourceCount - targetCount);
for (int i = sourceOffset + fromIndex; i <= max; i++) {
/* Look for first character. */
if (source[i] != first) {
while (++i <= max && source[i] != first)
;
}
/* Found first character, now look at the rest of v2 */
if (i <= max) {
int j = i + 1;
int end = j + targetCount - 1;
for (int k = targetOffset + 1; j < end && source[j] == target[k]; j++, k++)
;
if (j == end) {
/* Found whole string. */
return i - sourceOffset;
}
}
}
return -1;
}
}
LeetCode_28. Implement strStr()
标签:lan count The array targe character inter imp uri
原文地址:https://www.cnblogs.com/denggelin/p/11557870.html