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# LeetCode : 28. Implement strStr( )

2020-12-18
goodGid

## 28. Implement strStr( )

### Problem

``````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

``````Input: haystack = "hello", needle = "ll"
Output: 2
``````

### [1] Code (20. 12. 18)

``````class Solution {
public int strStr(String haystack, String needle) {
if (needle.equals("")) {
return 0;
}

if (!haystack.contains(needle)) {
return -1;
}

for (int i = 0; i < haystack.length(); i++) {
if (haystack.charAt(i) == needle.charAt(0)) {

boolean isContain = true;
for (int j = 0; j < needle.length(); j++) {
if (haystack.charAt(i + j) != needle.charAt(j)) {
isContain = false;
break;
}
}

if (isContain) {
return i;
}
}
}

return -1;
}
}
``````
• 하나하나씩 찾는다.

### [2] Code (20. 12. 18)

``````class Solution {
public int strStr(String haystack, String needle) {
int len = needle.length();

for (int i = 0; i < haystack.length() - len + 1; i++) {
String s = haystack.substring(i, i + len);

if (s.equals(needle)) {
return i;
}
}
return -1;
}
}
``````
• haystack을 needle의 length만큼 substring해서 비교하는 아이디어도 있었다.

### [3] Code (20. 12. 18)

``````class Solution {
public int strStr(String haystack, String needle) {
int idx = 0;

if (needle.length() == 0) {
return 0;
}

idx = haystack.indexOf(needle, 0);

return idx;
}
}
``````
• .indexOf( ) 메소드를 사용하면 해당 index를 바로 알 수 있다.