实现 strStr() 函数。

给定一个 haystack 字符串和一个 needle 字符串，在 haystack 字符串中找出 needle 字符串出现的第一个位置 (从0开始)。如果不存在，则返回  -1。

示例 1:

输入: haystack = "hello", needle = "ll"
输出: 2
示例 2:

输入: haystack = "aaaaa", needle = "bba"
输出: -1
说明:

当 needle 是空字符串时，我们应当返回什么值呢？这是一个在面试中很好的问题。

对于本题而言，当 needle 是空字符串时我们应当返回 0 。这与C语言的 strstr() 以及 Java的 indexOf() 定义相符。

解法一：

利用 Java String 的 indexOf 方法可以迅速求解。

    public int strStr(String haystack, String needle) {

        return haystack.indexOf(needle);
    }

indexOf 的源码如下：

    /**
     * Code shared by String and StringBuffer to do searches. The
     * source is the character array being searched, and the target
     * is the string being searched for.
     *
     * @param   source       the characters being searched.
     * @param   sourceOffset offset of the source string.
     * @param   sourceCount  count of the source string.
     * @param   target       the characters being searched for.
     * @param   targetOffset offset of the target string.
     * @param   targetCount  count of the target string.
     * @param   fromIndex    the index to begin searching from.
     */
    static int indexOf(char[] source, int sourceOffset, int sourceCount,
            char[] target, int targetOffset, int targetCount,
            int fromIndex) {
        if (fromIndex >= sourceCount) {
            return (targetCount == 0 ? sourceCount : -1);
        }
        if (fromIndex < 0) {
            fromIndex = 0;
        }
        if (targetCount == 0) {
            return fromIndex;
        }

        char first = target[targetOffset];
        int max = sourceOffset + (sourceCount - targetCount);

        for (int i = sourceOffset + fromIndex; i <= max; i++) {
            /* Look for first character. */
            if (source[i] != first) {
                while (++i <= max && source[i] != first);
            }

            /* Found first character, now look at the rest of v2 */
            if (i <= max) {
                int j = i + 1;
                int end = j + targetCount - 1;
                for (int k = targetOffset + 1; j < end && source[j]
                        == target[k]; j++, k++);

                if (j == end) {
                    /* Found whole string. */
                    return i - sourceOffset;
                }
            }
        }
        return -1;
    }