List removeAll详解

list.removeAll 会随着数量的增加，性能变得很差，原因为： list.contains 需要进行两次遍历

private boolean batchRemove(Collection<?> c, boolean complement) {
        final Object[] elementData = this.elementData;
        int r = 0, w = 0;
        boolean modified = false;
        try {
            for (; r < size; r++)
                if (c.contains(elementData[r]) == complement)
                    elementData[w++] = elementData[r];
        } finally {
            // Preserve behavioral compatibility with AbstractCollection,
            // even if c.contains() throws.
            if (r != size) {
                System.arraycopy(elementData, r,
                                 elementData, w,
                                 size - r);
                w += size - r;
            }
            if (w != size) {
                // clear to let GC do its work
                for (int i = w; i < size; i++)
                    elementData[i] = null;
                modCount += size - w;
                size = w;
                modified = true;
            }
        }
        return modified;
    }

　　

public boolean contains(Object o) {
        return indexOf(o) >= 0;
    }

    /**
     * Returns the index of the first occurrence of the specified element
     * in this list, or -1 if this list does not contain the element.
     * More formally, returns the lowest index <tt>i</tt> such that
     * <tt>(o==null ? get(i)==null : o.equals(get(i)))</tt>,
     * or -1 if there is no such index.
     */
    public int indexOf(Object o) {
        if (o == null) {
            for (int i = 0; i < size; i++)
                if (elementData[i]==null)
                    return i;
        } else {
            for (int i = 0; i < size; i++)
                if (o.equals(elementData[i]))
                    return i;
        }
        return -1;
    }

　　

解法：使用HashSet 和LinkedList 进行替换（HashSet保证元素唯一性，LinkedList提升插入删除性能）

public List<T> removeAll(List<T> source, List<T> destination) {
   List<T> result = new LinkedList<T>();
   Set<T> destinationSet = new HashSet<T>(destination);
   for(T t : source) {
       if (!destinationSet.contains(t)) {
	   result.add(t);
       }
   }
   return result;
}

　　

评述：主要是使用HashSet具有散列性质的contains 替换 需要进行两次遍历的contains