给定一个单链表 L:L0→L1→…→Ln-1→Ln ,
将其重新排列后变为: L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例 1:
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
这个题可以先用快慢指针找到中点保存,然后截断,将后半段存入栈中,然后遍历前半段,把栈中的元素插进链表,栈起到了翻转链表的作用。
C++源代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void reorderList(ListNode* head) {
if(!head) return;
ListNode *slow=head, *fast=head;
stack<ListNode*> s;
while(fast && fast->next){
slow = slow->next;
fast = fast->next->next;
}
fast = slow->next;
slow->next = NULL;
slow = head;
while(fast){
s.push(fast);
fast = fast->next;
}
while(!s.empty()){
ListNode *tmp = slow->next;
slow->next = s.top();
slow->next->next = tmp;
slow = tmp;
s.pop();
}
}
};
python3源代码:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reorderList(self, head: ListNode) -> None:
"""
Do not return anything, modify head in-place instead.
"""
if head==None:
return
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
fast = slow.next
slow.next = None
slow = head
s = []
while fast:
s.append(fast)
fast = fast.next
while len(s)!=0:
tmp = slow.next
slow.next = s.pop()
slow.next.next = tmp
slow = tmp