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<h1 id="双单向链表"><a class="header" href="#双单向链表">双单向链表</a></h1>
<p>在之前的双向链表章节中，我们一度非常纠结，原因来自同样纠结成一团的所有权依赖。还有一个重要原因就是：先入为主的链表定义。</p>
<p>谁说所有的链接一定要一个方向呢？这里一起来尝试下新的东东：链表的其中一半朝左，另一半朝右。</p>
<p>新规矩( 老规矩是创建文件 )，创建一个新的模块:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>// lib.rs
// ...
pub mod silly1; // NEW!
<span class="boring">}</span></code></pre></pre>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>// silly1.rs
use crate::second::List as Stack;
struct List&lt;T&gt; {
left: Stack&lt;T&gt;,
right: Stack&lt;T&gt;,
}
<span class="boring">}</span></code></pre></pre>
<p>这里将之前的 <code>List</code> 引入进来，并重命名为 <code>Stack</code>，接着，创建一个新的链表。现在既可以向左增长又可以向右增长。</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub struct Stack&lt;T&gt; {
head: Link&lt;T&gt;,
}
type Link&lt;T&gt; = Option&lt;Box&lt;Node&lt;T&gt;&gt;&gt;;
struct Node&lt;T&gt; {
elem: T,
next: Link&lt;T&gt;,
}
impl&lt;T&gt; Stack&lt;T&gt; {
pub fn new() -&gt; Self {
Stack { head: None }
}
pub fn push(&amp;mut self, elem: T) {
let new_node = Box::new(Node {
elem: elem,
next: self.head.take(),
});
self.head = Some(new_node);
}
pub fn pop(&amp;mut self) -&gt; Option&lt;T&gt; {
self.head.take().map(|node| {
let node = *node;
self.head = node.next;
node.elem
})
}
pub fn peek(&amp;self) -&gt; Option&lt;&amp;T&gt; {
self.head.as_ref().map(|node| {
&amp;node.elem
})
}
pub fn peek_mut(&amp;mut self) -&gt; Option&lt;&amp;mut T&gt; {
self.head.as_mut().map(|node| {
&amp;mut node.elem
})
}
}
impl&lt;T&gt; Drop for Stack&lt;T&gt; {
fn drop(&amp;mut self) {
let mut cur_link = self.head.take();
while let Some(mut boxed_node) = cur_link {
cur_link = boxed_node.next.take();
}
}
}
<span class="boring">}</span></code></pre></pre>
<p>稍微修改下 <code>push</code> 和 <code>pop</code>：</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub fn push(&amp;mut self, elem: T) {
let new_node = Box::new(Node {
elem: elem,
next: None,
});
self.push_node(new_node);
}
fn push_node(&amp;mut self, mut node: Box&lt;Node&lt;T&gt;&gt;) {
node.next = self.head.take();
self.head = Some(node);
}
pub fn pop(&amp;mut self) -&gt; Option&lt;T&gt; {
self.pop_node().map(|node| {
node.elem
})
}
fn pop_node(&amp;mut self) -&gt; Option&lt;Box&lt;Node&lt;T&gt;&gt;&gt; {
self.head.take().map(|mut node| {
self.head = node.next.take();
node
})
}
<span class="boring">}</span></code></pre></pre>
<p>现在可以开始构造新的链表:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub struct List&lt;T&gt; {
left: Stack&lt;T&gt;,
right: Stack&lt;T&gt;,
}
impl&lt;T&gt; List&lt;T&gt; {
fn new() -&gt; Self {
List { left: Stack::new(), right: Stack::new() }
}
}
<span class="boring">}</span></code></pre></pre>
<p>当然，还有一大堆左左右右类型的操作:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub fn push_left(&amp;mut self, elem: T) { self.left.push(elem) }
pub fn push_right(&amp;mut self, elem: T) { self.right.push(elem) }
pub fn pop_left(&amp;mut self) -&gt; Option&lt;T&gt; { self.left.pop() }
pub fn pop_right(&amp;mut self) -&gt; Option&lt;T&gt; { self.right.pop() }
pub fn peek_left(&amp;self) -&gt; Option&lt;&amp;T&gt; { self.left.peek() }
pub fn peek_right(&amp;self) -&gt; Option&lt;&amp;T&gt; { self.right.peek() }
pub fn peek_left_mut(&amp;mut self) -&gt; Option&lt;&amp;mut T&gt; { self.left.peek_mut() }
pub fn peek_right_mut(&amp;mut self) -&gt; Option&lt;&amp;mut T&gt; { self.right.peek_mut() }
<span class="boring">}</span></code></pre></pre>
<p>其中最有趣的是：还可以来回闲逛了。</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub fn go_left(&amp;mut self) -&gt; bool {
self.left.pop_node().map(|node| {
self.right.push_node(node);
}).is_some()
}
pub fn go_right(&amp;mut self) -&gt; bool {
self.right.pop_node().map(|node| {
self.left.push_node(node);
}).is_some()
}
<span class="boring">}</span></code></pre></pre>
<p>这里返回 <code>bool</code> 是为了告诉调用者我们是否成功的移动。最后，再来测试下：</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>#[cfg(test)]
mod test {
use super::List;
#[test]
fn walk_aboot() {
let mut list = List::new(); // [_]
list.push_left(0); // [0,_]
list.push_right(1); // [0, _, 1]
assert_eq!(list.peek_left(), Some(&amp;0));
assert_eq!(list.peek_right(), Some(&amp;1));
list.push_left(2); // [0, 2, _, 1]
list.push_left(3); // [0, 2, 3, _, 1]
list.push_right(4); // [0, 2, 3, _, 4, 1]
while list.go_left() {} // [_, 0, 2, 3, 4, 1]
assert_eq!(list.pop_left(), None);
assert_eq!(list.pop_right(), Some(0)); // [_, 2, 3, 4, 1]
assert_eq!(list.pop_right(), Some(2)); // [_, 3, 4, 1]
list.push_left(5); // [5, _, 3, 4, 1]
assert_eq!(list.pop_right(), Some(3)); // [5, _, 4, 1]
assert_eq!(list.pop_left(), Some(5)); // [_, 4, 1]
assert_eq!(list.pop_right(), Some(4)); // [_, 1]
assert_eq!(list.pop_right(), Some(1)); // [_]
assert_eq!(list.pop_right(), None);
assert_eq!(list.pop_left(), None);
}
}
<span class="boring">}</span></code></pre></pre>
<pre><code class="language-shell">$ cargo test
Running target/debug/lists-5c71138492ad4b4a
running 16 tests
test fifth::test::into_iter ... ok
test fifth::test::basics ... ok
test fifth::test::iter ... ok
test fifth::test::iter_mut ... ok
test fourth::test::into_iter ... ok
test fourth::test::basics ... ok
test fourth::test::peek ... ok
test first::test::basics ... ok
test second::test::into_iter ... ok
test second::test::basics ... ok
test second::test::iter ... ok
test second::test::iter_mut ... ok
test third::test::basics ... ok
test third::test::iter ... ok
test second::test::peek ... ok
test silly1::test::walk_aboot ... ok
test result: ok. 16 passed; 0 failed; 0 ignored; 0 measured
</code></pre>
<p>上上下下，左左右右，BABA，哦耶，这个链表无敌了！</p>
<p>以上是一个非常典型的<ruby>手指型数据结构<rt>finger data structure</rt></ruby>，在其中维护一个手指，然后操作所需的时间与手指的距离成正比。</p>
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