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# 28. Implement strStr() | ||
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Implement strStr(). | ||
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Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. | ||
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**Clarification:** | ||
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What should we return when needle is an empty string? This is a great question to ask during an interview. | ||
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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(). | ||
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**Example 1:** | ||
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``` | ||
Input: haystack = "hello", needle = "ll" | ||
Output: 2 | ||
``` | ||
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**Example 2:** | ||
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``` | ||
Input: haystack = "aaaaa", needle = "bba" | ||
Output: -1 | ||
``` | ||
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**Example 3:** | ||
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``` | ||
Input: haystack = "", needle = "" | ||
Output: 0 | ||
``` | ||
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**Constraints:** | ||
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* 0 <= haystack.length, needle.length <= 5 * 104 | ||
* haystack and needle consist of only lower-case English characters. |
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/** | ||
* @param {string} haystack | ||
* @param {string} needle | ||
* @return {number} | ||
*/ | ||
var strStr = function(haystack, needle) { | ||
if (needle === "") return 0; | ||
if (needle.length > haystack.length) return -1; | ||
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let haystackPointer = 0; | ||
let needlePointer = 0; | ||
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while (haystackPointer < haystack.length && needlePointer < needle.length) { | ||
if (haystack[haystackPointer] === needle[needlePointer]) { | ||
haystackPointer++; | ||
needlePointer++; | ||
} else { | ||
haystackPointer = haystackPointer + 1 - needlePointer; | ||
needlePointer = 0; | ||
} | ||
} | ||
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if (needlePointer === needle.length) return haystackPointer - needle.length; | ||
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return -1; | ||
}; |