碰撞 collision

鏈表解決

https://openjdk.org/projects/jdk7/

https://jdk.java.net/java-se-ri/7

https://www.oracle.com/java/technologies/javase/javase7-archive-downloads.html

1、負(fù)數(shù)mod為負(fù)數(shù)

2、較慢

mod本質(zhì)是除法，相較于位運(yùn)算

實(shí)際jdk1.7實(shí)現(xiàn)：

hash & ( lenght - 1)

static int indexFor(int h, int length) {
    // assert Integer.bitCount(length) == 1 : "length must be a non-zero power of 2";
    return h & (length-1);
}

圖中老師有筆誤

32 = DEC 10,0000

32 -1 = DEC 1,1111

假如不是 a power of two，length - 1 就不是全1了

彈幕參考：相當(dāng)于就是用按位與的快速運(yùn)算代替了非常笨重的取余運(yùn)算

（很取巧）

/**
 * Retrieve object hash code and applies a supplemental hash function to the
 * result hash, which defends against poor quality hash functions.  This is
 * critical because HashMap uses power-of-two length hash tables, that
 * otherwise encounter collisions for hashCodes that do not differ
 * in lower bits. Note: Null keys always map to hash 0, thus index 0.
 */
final int hash(Object k) {
    int h = hashSeed;
    if (0 != h && k instanceof String) {
        return sun.misc.Hashing.stringHash32((String) k);
    }

    h ^= k.hashCode();

    // This function ensures that hashCodes that differ only by
    // constant multiples at each bit position have a bounded
    // number of collisions (approximately 8 at default load factor).
    h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);
}