How collections.deque works?
前言：在 Python 生态中，我们经常使用 collections.deque 来实现栈、队列这些只需要进行头尾操作的数据结构，它的 append/pop 操作都是 O(1) 时间复杂度。list 的 pop(0) 的时间复杂度是 O(n)， 在这个场景中，它的效率没有 deque 高。那 deque 内部是怎样实现的呢？ 我从 GitHub 上挖出了 CPython collections 模块的第二个 commit 的源码。
dequeobject 对象定义
注释写得优雅了，无法进行更加精简的总结。
/* The block length may be set to any number over 1. Larger numbers
* reduce the number of calls to the memory allocator but take more
* memory. Ideally, BLOCKLEN should be set with an eye to the
* length of a cache line.
*/
#define BLOCKLEN 62
#define CENTER ((BLOCKLEN - 1) / 2)
/* A `dequeobject` is composed of a doubly-linked list of `block` nodes.
* This list is not circular (the leftmost block has leftlink==NULL,
* and the rightmost block has rightlink==NULL). A deque d's first
* element is at d.leftblock[leftindex] and its last element is at
* d.rightblock[rightindex]; note that, unlike as for Python slice
* indices, these indices are inclusive on both ends. By being inclusive
* on both ends, algorithms for left and right operations become
* symmetrical which simplifies the design.
*
* The list of blocks is never empty, so d.leftblock and d.rightblock
* are never equal to NULL.
*
* The indices, d.leftindex and d.rightindex are always in the range
* 0 <= index < BLOCKLEN.
* Their exact relationship is:
* (d.leftindex + d.len - 1) % BLOCKLEN == d.rightindex.
*
* Empty deques have d.len == 0; d.leftblock==d.rightblock;
* d.leftindex == CENTER+1; and d.rightindex == CENTER.
* Checking for d.len == 0 is the intended way to see whether d is empty.
*
* Whenever d.leftblock == d.rightblock,
* d.leftindex + d.len - 1 == d.rightindex.
*
* However, when d.leftblock != d.rightblock, d.leftindex and d.rightindex
* become indices into distinct blocks and either may be larger than the
* other.
*/
typedef struct BLOCK {
struct BLOCK *leftlink;
struct BLOCK *rightlink;
PyObject *data[BLOCKLEN];
} block;
typedef struct {
PyObject_HEAD
block *leftblock;
block *rightblock;
int leftindex; /* in range(BLOCKLEN) */
int rightindex; /* in range(BLOCKLEN) */
int len;
long state; /* incremented whenever the indices move */
PyObject *weakreflist; /* List of weak references */
} dequeobject;下面是我为 Block 结构体画的一个图
+----------------------------------------+
| data: 62 objects |
+----------+ | | +-----------+
| leftlink |---| | ... | Obj1 | Obj2 | Obj3 | ... | |---| rightlink |
+----------+ | 30 31 32 | +-----------+
+----------------------------------------+创建一个 block
static block *
newblock(block *leftlink, block *rightlink, int len) {
block *b;
/* To prevent len from overflowing INT_MAX on 64-bit machines, we
* refuse to allocate new blocks if the current len is dangerously
* close. There is some extra margin to prevent spurious arithmetic
* overflows at various places. The following check ensures that
* the blocks allocated to the deque, in the worst case, can only
* have INT_MAX-2 entries in total.
*/
if (len >= INT_MAX - 2*BLOCKLEN) {
PyErr_SetString(PyExc_OverflowError,
"cannot add more blocks to the deque");
return NULL;
}
b = PyMem_Malloc(sizeof(block));
if (b == NULL) {
PyErr_NoMemory();
return NULL;
}
b->leftlink = leftlink;
b->rightlink = rightlink;
return b;
}创建一个 dequeobject
- 创建一个 block
- 实例化一个 dequeobject Python 对象（这一块的内在逻辑目前我也不太懂）
- leftblock 和 rightblock 指针都指向这个 block
- leftindex 是 CENTER+1，rightindex 是 CENTER
- 初始化其他一些属性， len state 等
这个第一步和第四步都有点意思，第一步创建一个 block，也就是说， deque 对象创建的时候，就预先分配了一块内存。第四步隐约告诉我们， 当元素来的时候，它先会被放在中间，然后逐渐往头和尾散开。
static PyObject *
deque_new(PyTypeObject *type, PyObject *args, PyObject *kwds)
{
dequeobject *deque;
block *b;
if (type == &deque_type && !_PyArg_NoKeywords("deque()", kwds))
return NULL;
/* create dequeobject structure */
deque = (dequeobject *)type->tp_alloc(type, 0);
if (deque == NULL)
return NULL;
b = newblock(NULL, NULL, 0);
if (b == NULL) {
Py_DECREF(deque);
return NULL;
}
assert(BLOCKLEN >= 2);
deque->leftblock = b;
deque->rightblock = b;
deque->leftindex = CENTER + 1;
deque->rightindex = CENTER;
deque->len = 0;
deque->state = 0;
deque->weakreflist = NULL;
return (PyObject *)deque;
}deque.append 实现
步骤：
- 如果 rightblock 可以容纳更多的元素，则放在 rightblock 中
- 如果不能，就新建一个 block，然后更新若干指针，将元素放在更新后的 rightblock 中
static PyObject *
deque_append(dequeobject *deque, PyObject *item)
{
deque->state++;
if (deque->rightindex == BLOCKLEN-1) {
block *b = newblock(deque->rightblock, NULL, deque->len);
if (b == NULL)
return NULL;
assert(deque->rightblock->rightlink == NULL);
deque->rightblock->rightlink = b;
deque->rightblock = b;
deque->rightindex = -1;
}
Py_INCREF(item);
deque->len++;
deque->rightindex++;
deque->rightblock->data[deque->rightindex] = item;
Py_RETURN_NONE;
}看了 append 实现后，我们可以自行脑补一下 pop 和 popleft 的实现。
小结
deque 内部将一组内存块组织成双向链表的形式，每个内存块可以看成一个 Python 对象的数组， 这个数组与普通数据不同，它是从数组中部往头尾两边填充数据，而平常所见数组大都是从头往后。 得益于 deque 这样的结构，它的 pop/popleft/append/appendleft 四种操作的时间复杂度均是 O(1), 用它来实现队列、栈数据结构会非常方便和高效。但也正因为这样的设计， 它不能像数组那样通过 index 来访问、移除元素。链表 + 数组、或者链表 + 字典 这样的设计在实践中有很广泛的应用，比如 LRUCache, LFUCache，有兴趣的同鞋可以继续探索。
- PS1: LRUCache 在面试中不要太常见
- PS2: 出 LFUCache 题的面试官都是变态
- PS3: 头图来自 quora ，图文不怎么有关系列