用于在链表中从开始到结束交换第 k 个节点的 Javascript 程序
给定一个单链表，从开始交换第 k 个节点，从结束交换第 k 个节点。不允许数据交换，只应更改指针。在链表数据部分很大的许多情况下，这个要求可能是合乎逻辑的(例如，学生详细信息行名称、行号、地址，..等等)。指针总是固定的(大多数编译器为 4 字节)。 T3】例:
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 2
Output: 1 -> 4 -> 3 -> 2 -> 5
Explanation: The 2nd node from 1st is 2 and
2nd node from last is 4, so swap them.
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 5
Output: 5 -> 2 -> 3 -> 4 -> 1
Explanation: The 5th node from 1st is 5 and
5th node from last is 1, so swap them.
插图:
方法:思路很简单，从开始找第 k 个节点，最后第 k 个节点从开始就是第 n-k+1 个节点。交换两个节点。 不过也有一些角落案例，必须要处理
- y 紧挨着 X
- x 紧挨着 Y
- x 和 Y 是一样的
- x 和 Y 不存在(k 大于链表中的节点数)
下面是上述方法的实现。
java 描述语言
<script>
// A javascript program to swap kth
// node from the beginning with
// kth node from the end
class Node
{
constructor(val)
{
this.data = val;
this.next = null;
}
}
var head;
/* Utility function to insert a node
at the beginning */
function push(new_data)
{
new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
/* Utility function for displaying
linked list */
function printList()
{
node = head;
while (node != null)
{
document.write(node.data + " ");
node = node.next;
}
document.write("");
}
/* Utility function for calculating length
of linked list */
function countNodes()
{
var count = 0;
s = head;
while (s != null)
{
count++;
s = s.next;
}
return count;
}
/* Function for swapping kth nodes from
both ends of linked list */
function swapKth(k)
{
// Count nodes in linked list
var n = countNodes();
// Check if k is valid
if (n < k)
return;
// If x (kth node from start) and
// y(kth node from end) are same
if (2 * k - 1 == n)
return;
// Find the kth node from beginning
// of linked list. We also find previous
// of kth node because we need to update
// next pointer of the previous.
x = head;
x_prev = null;
for (i = 1; i < k; i++)
{
x_prev = x;
x = x.next;
}
// Similarly, find the kth node from end
// and its previous. kth node from end
// is (n-k+1)th node from beginning
y = head;
y_prev = null;
for (i = 1; i < n - k + 1; i++)
{
y_prev = y;
y = y.next;
}
// If x_prev exists, then new next of it
// will be y. Consider the case when y->next
// is x, in this case, x_prev and y are same.
// So the statement "x_prev->next = y" creates
// a self loop. This self loop will be broken
// when we change y->next.
if (x_prev != null)
x_prev.next = y;
// Same thing applies to y_prev
if (y_prev != null)
y_prev.next = x;
// Swap next pointers of x and y. These
// statements also break self loop if
// x->next is y or y->next is x
temp = x.next;
x.next = y.next;
y.next = temp;
// Change head pointers when k is 1 or n
if (k == 1)
head = y;
if (k == n)
head = x;
}
// Driver code
for (let i = 8; i >= 1; i--)
push(i);
document.write(
"Original linked list: <br/>");
printList();
document.write("<br/>");
for (let i = 1; i < 9; i++)
{
swapKth(i);
document.write(
"Modified List for k = " + i + "<br/>");
printList();
document.write("<br/>");
}
// This code is contributed by gauravrajput1
</script>
输出:
Original Linked List: 1 2 3 4 5 6 7 8
Modified List for k = 1
8 2 3 4 5 6 7 1
Modified List for k = 2
8 7 3 4 5 6 2 1
Modified List for k = 3
8 7 6 4 5 3 2 1
Modified List for k = 4
8 7 6 5 4 3 2 1
Modified List for k = 5
8 7 6 4 5 3 2 1
Modified List for k = 6
8 7 3 4 5 6 2 1
Modified List for k = 7
8 2 3 4 5 6 7 1
Modified List for k = 8
1 2 3 4 5 6 7 8
复杂度分析:
- 时间复杂度: O(n)，其中 n 为列表长度。 需要遍历列表一次。
- 辅助空间: O(1)。 不需要额外空间。
更多详情请参考完整文章在链表中从开始到结束交换第 k 个节点！
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