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: 2Example 2:
Input: haystack = “aaaaa”, needle = “bba”
Output: -1Clarification:
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().
strStr() implementation in C++
The implementation should consider the corner cases, when the needle is empty, the result should be always 0. The following implementation of strStr() method in C++ has a time complexity O(MN) where M and N are the lengths of the input haystack and needle string.
The outer loop should be from 0 to the length of M-N+1. And the inner loop checks if the substring of haystack matches the needle. This is not the optimial solution as there will be a KMP algorithm which requires a pre-computation.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | class Solution { public: int strStr(string haystack, string needle) { for (int i = 0; i + needle.size() - 1 < haystack.size(); ++ i) { bool ok = true; for (int j = 0; j < needle.size(); ++ j) { if (haystack[i + j] != needle[j]) { ok = false; break; } } if (ok) return i; } return (needle.size() == 0) ? 0 : -1; } }; |
class Solution { public: int strStr(string haystack, string needle) { for (int i = 0; i + needle.size() - 1 < haystack.size(); ++ i) { bool ok = true; for (int j = 0; j < needle.size(); ++ j) { if (haystack[i + j] != needle[j]) { ok = false; break; } } if (ok) return i; } return (needle.size() == 0) ? 0 : -1; } };
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