题目如图：
private static int strStr(String haystack, String needle) {if (haystack == null || needle == null) {return -1;}int needleLength = needle.length();if (needleLength == 0) {return 0;}int stackLength = haystack.length();if (needleLength > stackLength) {return -1;}int i = 0, j = 0, count = 0;// i < stackLength - (needleLength - j - 1) 目的是为了减少查找次数// 例如 A = "aabbccdef" B = "def" 当 j = 0 的时候最多只能到A串的'd',后面的“ef”没必要查找// 当 j = 1 的时候 最多只能到 A 中的 'e' 相当于动态限制for (; j < needleLength && i < stackLength - (needleLength - j - 1); i++, j++) {while (haystack.charAt(i) != needle.charAt(j) && i < stackLength - (needleLength - j - 1)) {// 全部遍历了还没找到 只针对于 needleLength = 1 或者两个字符串长度相等if (i == stackLength - 1 || needleLength == stackLength) {return -1;}if (count == 0) {// 没有找到过，就一直找下去++i;} else {// 找到过，但是中间某个不匹配了，要重新找 即 j 指针指向 0// 例如 abccbcdd 和 bcd 之前匹配到 bc count = 2 // 然后 c 和 d 不匹配，所以需要重新匹配，i 从之前的 index 即 (i - count） 再加 1 位出发就好j = 0;i = (i - count) + 1;count = 0;}}if (++count == needleLength && i < stackLength - (needleLength - j - 1)) {return i - count + 1;}}return -1;
}
复制代码看了第一名的代码，愈发觉得jdk作者的强大?
