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//! A singly-linked list is headed by a single forward pointer. The elements
//! are singly-linked for minimum space and pointer manipulation overhead at
//! the expense of O(n) removal for arbitrary elements. New elements can be
//! added to the list after an existing element or at the head of the list.
//!
//! A singly-linked list may only be traversed in the forward direction.
//!
//! Singly-linked lists are useful under these conditions:
//! * Ability to preallocate elements / requirement of infallibility for
//! insertion.
//! * Ability to allocate elements intrusively along with other data.
//! * Homogenous elements.
const std = @import("std.zig");
const debug = std.debug;
const assert = debug.assert;
const testing = std.testing;
const SinglyLinkedList = @This();
first: ?*Node = null,
/// This struct contains only a next pointer and not any data payload. The
/// intended usage is to embed it intrusively into another data structure and
/// access the data with `@fieldParentPtr`.
pub const Node = struct {
next: ?*Node = null,
pub fn insertAfter(node: *Node, new_node: *Node) void {
new_node.next = node.next;
node.next = new_node;
}
/// Remove the node after the one provided, returning it.
pub fn removeNext(node: *Node) ?*Node {
const next_node = node.next orelse return null;
node.next = next_node.next;
return next_node;
}
/// Iterate over the singly-linked list from this node, until the final
/// node is found.
///
/// This operation is O(N). Instead of calling this function, consider
/// using a different data structure.
pub fn findLast(node: *Node) *Node {
var it = node;
while (true) {
it = it.next orelse return it;
}
}
/// Iterate over each next node, returning the count of all nodes except
/// the starting one.
///
/// This operation is O(N). Instead of calling this function, consider
/// using a different data structure.
pub fn countChildren(node: *const Node) usize {
var count: usize = 0;
var it: ?*const Node = node.next;
while (it) |n| : (it = n.next) {
count += 1;
}
return count;
}
/// Reverse the list starting from this node in-place.
///
/// This operation is O(N). Instead of calling this function, consider
/// using a different data structure.
pub fn reverse(indirect: *?*Node) void {
if (indirect.* == null) {
return;
}
var current: *Node = indirect.*.?;
while (current.next) |next| {
current.next = next.next;
next.next = indirect.*;
indirect.* = next;
}
}
};
pub fn prepend(list: *SinglyLinkedList, new_node: *Node) void {
new_node.next = list.first;
list.first = new_node;
}
pub fn remove(list: *SinglyLinkedList, node: *Node) void {
if (list.first == node) {
list.first = node.next;
} else {
var current_elm = list.first.?;
while (current_elm.next != node) {
current_elm = current_elm.next.?;
}
current_elm.next = node.next;
}
}
/// Remove and return the first node in the list.
pub fn popFirst(list: *SinglyLinkedList) ?*Node {
const first = list.first orelse return null;
list.first = first.next;
return first;
}
/// Iterate over all nodes, returning the count.
///
/// This operation is O(N). Consider tracking the length separately rather than
/// computing it.
pub fn len(list: SinglyLinkedList) usize {
if (list.first) |n| {
return 1 + n.countChildren();
} else {
return 0;
}
}
test "basics" {
const L = struct {
data: u32,
node: SinglyLinkedList.Node = .{},
};
var list: SinglyLinkedList = .{};
try testing.expect(list.len() == 0);
var one: L = .{ .data = 1 };
var two: L = .{ .data = 2 };
var three: L = .{ .data = 3 };
var four: L = .{ .data = 4 };
var five: L = .{ .data = 5 };
list.prepend(&two.node); // {2}
two.node.insertAfter(&five.node); // {2, 5}
list.prepend(&one.node); // {1, 2, 5}
two.node.insertAfter(&three.node); // {1, 2, 3, 5}
three.node.insertAfter(&four.node); // {1, 2, 3, 4, 5}
try testing.expect(list.len() == 5);
// Traverse forwards.
{
var it = list.first;
var index: u32 = 1;
while (it) |node| : (it = node.next) {
const l: *L = @fieldParentPtr("node", node);
try testing.expect(l.data == index);
index += 1;
}
}
_ = list.popFirst(); // {2, 3, 4, 5}
_ = list.remove(&five.node); // {2, 3, 4}
_ = two.node.removeNext(); // {2, 4}
try testing.expect(@as(*L, @fieldParentPtr("node", list.first.?)).data == 2);
try testing.expect(@as(*L, @fieldParentPtr("node", list.first.?.next.?)).data == 4);
try testing.expect(list.first.?.next.?.next == null);
SinglyLinkedList.Node.reverse(&list.first);
try testing.expect(@as(*L, @fieldParentPtr("node", list.first.?)).data == 4);
try testing.expect(@as(*L, @fieldParentPtr("node", list.first.?.next.?)).data == 2);
try testing.expect(list.first.?.next.?.next == null);
}
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