Find the Index of the First Occurrence in a String
<p>Given two strings <code>needle</code> and <code>haystack</code>, return the index of the first occurrence of <code>needle</code> in <code>haystack</code>, or <code>-1</code> if <code>needle</code> is not part of <code>haystack</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> haystack = &quot;sadbutsad&quot;, needle = &quot;sad&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> &quot;sad&quot; occurs at index 0 and 6.
The first occurrence is at index 0, so we return 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> haystack = &quot;leetcode&quot;, needle = &quot;leeto&quot;
<strong>Output:</strong> -1
<strong>Explanation:</strong> &quot;leeto&quot; did not occur in &quot;leetcode&quot;, so we return -1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= haystack.length, needle.length &lt;= 10<sup>4</sup></code></li>
	<li><code>haystack</code> and <code>needle</code> consist of only lowercase English characters.</li>
</ul>
Two Pointers; String; String Matching
class Solution {
private:
    vector<int> Next(string str) {
        vector<int> n(str.length());
        n[0] = -1;
        int i = 0, pre = -1;
        int len = str.length();
        while (i < len) {
            while (pre >= 0 && str[i] != str[pre])
                pre = n[pre];
            ++i, ++pre;
            if (i >= len)
                break;
            if (str[i] == str[pre])
                n[i] = n[pre];
            else
                n[i] = pre;
        }
        return n;
    }

public:
    int strStr(string haystack, string needle) {
        if (0 == needle.length())
            return 0;

        vector<int> n(Next(needle));

        int len = haystack.length() - needle.length() + 1;
        for (int i = 0; i < len; ++i) {
            int j = 0, k = i;
            while (j < needle.length() && k < haystack.length()) {
                if (haystack[k] != needle[j]) {
                    if (n[j] >= 0) {
                        j = n[j];
                        continue;
                    } else
                        break;
                }
                ++k, ++j;
            }
            if (j >= needle.length())
                return k - j;
        }

        return -1;
    }
};