题目描述
Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…
You must do this in-place without altering the nodes' values.
For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}
解题思路
(1)快慢指针找到中间节点
(2)拆分链表,将后半部分链表反转
(3)合并两部分链表
代码示例
class Solution {
public:
void reorderList(ListNode *head) {
if (head == NULL || head->next == NULL) {
return ;
}
// 快满指针找到中间节点,(左右部分相等长度,或左边多一个)
ListNode *q = head;
while (q->next && q->next->next) {
p = p->next;
q = q->next->next;
}
// 拆分链表,并反转中间节点之后的链表(反复熟练该过程)
p->next = NULL;
ListNode *left = head;
ListNode *pre = NULL;
while (right) {
ListNode *temp = right->next;
right->next = pre;
pre = right;
right = temp;
}
// 合并两个链表
while (left && right) {
ListNode *temp1 = left->next;
ListNode *temp2 = right->next;
left->next = right;
right->next = temp1;
left = temp1;
right = temp2;
}
}
};