如果我定义了一个项目列表,并且我想更改或更改所选项目之一在列表中的顺序,我该如何实现?
函数的定义应如下:
items* shift_back(item* head, int i){
....
....
}
假设这是您的清单:
+-----+ +-----+ +-----+ +-----+ +-----+
|11| -|->|22| -|->|33| -|->|44| -|->|55| -|->nullptr
+-----+ +-----+ +-----+ +-----+ +-----+
1 2 3 4 5 [k represents position of a node]
如果
k
的值为 2
(即向上移动 2
nd 节点),那么您只需要有关其前一个节点(即第一个节点)和向上移动 2
nd 节点后的信息,它将成为第一个节点。
[k - 1] -> [k] -> [k + 1] -> [k + 1] -> .....
------- |
| +- This node next pointer need to set to [k - 1]
+- This node next pointer need to set to [k + 1]
// after moving up kth position node
[k] -> [k - 1] -> [k + 1] -> [k + 1] -> .....
example [k = 2]:
+-----+ +-----+ +-----+ +-----+ +-----+
|22| -|->|11| -|->|33| -|->|44| -|->|55| -|->nullptr
+-----+ +-----+ +-----+ +-----+ +-----+
如果
k
的值大于2
(2
nd节点之后的任何节点),则需要有关位置k
处节点的前两个节点的信息。
[k - 3] -> [k - 2] -> [k - 1] -> [k] -> [k + 1] -> [k + 1] -> .....
------- ------- |
| | +- This node next pointer need to set to [k - 1]
| +- This node next pointer need to set to [k + 1]
+- This node next pointer need to set to [k]
// after moving up kth position node up
[k - 3] -> [k - 2] -> [k] -> [k - 1] -> [k + 1] -> [k + 1] -> .....
example [k = 4]:
+-----+ +-----+ +-----+ +-----+ +-----+
|11| -|->|22| -|->|44| -|->|33| -|->|55| -|->nullptr
+-----+ +-----+ +-----+ +-----+ +-----+
如果
k
的值小于或等于1
或超出列表长度的任何值,则列表应保持不变。
牢记以上几点,你可以做到:
consumer* move_up (consumer* head, int k) {
consumer* ll = head;
consumer* prev = nullptr;
for (int i = 1; (i < k - 1) && (ll != nullptr); ++i) {
prev = ll;
ll = ll->next;
}
if ((k > 1) && (ll != nullptr) && (ll->next != nullptr)) {
consumer* x = ll->next;
ll->next = ll->next->next;
x->next = ll;
ll = x;
}
if (prev != nullptr) {
prev->next = ll;
ll = head;
}
return ll;
}
一些建议: