[leetcode 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.

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.

문제 풀이:

c언어에 문자열을 검색하는 strstr라는 함수가 있다는 걸 알게 되었다.

브루트포스로 풀었다.

class Solution {
public:
    int strStr(string haystack, string needle) {
        
        int m = needle.length();
        int n = haystack.length();
        if(m==0){
            return 0;
        }
        
        if(n<m){
            return -1;
        }
        
        int result = -1;
        for(int i=0; i<n; i++){
            if(haystack[i] == needle[0]){
                result = i;
                for(int j=1; j<m; j++){
                    if(haystack[i+j] != needle[j]){
                        result = -1;
                        break;
                    }
                }
                
                if(result > -1){
                    return result;
                }
            }
        }
        
        return result;
        
    }
};

 

++ 더 빠른 코드

class Solution {
public:
    int strStr(string haystack, string needle) {
        
        int m = needle.length();
        int n = haystack.length();
        if(m==0){
            return 0;
        }
        
        if(n<m){
            return -1;
        }
        
        int result = -1;
        for(int i=0; i<=n-m; i++){
            result = i;
            for(int j=0; j<m; j++){
                if(haystack[i+j] != needle[j]){
                    result = -1;
                    break;
                }
            }
             
            if(result > -1){
               return i;
            }
        }
        return result;
    }
};