给定一个单链表L: L0→L1→…→Ln-1→Ln,
重新排列后为:L0→Ln→L1→Ln-1→L2→Ln-2→…
必须在不改变节点值的情况下进行原地操作。
样例 给出链表 1->2->3->4->null,重新排列后为1->4->2->3->null。
算法很原始,先找到链表的中间节点,然后将后半部分提出来,把后半部分反转,然后将前半部分与反转了的后半部分合并起来,合并的时候注意交叉合并就可以了
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
*/
public class Solution {
/**
* @param head: The head of linked list.
* @return: void
*/
private ListNode reverse(ListNode head) {
ListNode newHead = null;
while (head != null) {
ListNode temp = head.next;
head.next = newHead;
newHead = head;
head = temp;
}
return newHead;
}
private void merge(ListNode head1, ListNode head2) {
int index = 0;
ListNode dummy = new ListNode(0);
while (head1 != null && head2 != null) {
if (index % 2 == 0) {
dummy.next = head1;
head1 = head1.next;
} else {
dummy.next = head2;
head2 = head2.next;
}
dummy = dummy.next;
index ++;
}
if (head1 != null) {
dummy.next = head1;
} else {
dummy.next = head2;
}
}
private ListNode findMiddle(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
public void reorderList(ListNode head) {
if (head == null || head.next == null) {
return;
}
ListNode mid = findMiddle(head);
ListNode tail = reverse(mid.next);
mid.next = null;
merge(head, tail);
}
}