LeetCode143:Reorder List
题目:
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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,先利用快慢指针找到链表中间节点
2,将链表后半部分进行反转
3,将链表前半部分与反转后的后半部分进行合并
实现代码:
#include <iostream> using namespace std; /* Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}. */ struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; void addNode(ListNode* &head, int val) { ListNode *node = new ListNode(val); if(head == NULL) { head = node; } else { node->next = head; head = node; } } void printList(ListNode *head) { while(head) { cout<<head->val<<" "; head = head->next; } } class Solution { public: void reorderList(ListNode *head) { if(head == NULL || head->next == NULL) return ; ListNode *quick = head; ListNode *slow = head; while(quick->next &&quick->next->next)//采用快慢指针查找链表中间节点,快指针走两步,慢指针走一步 { quick = quick->next->next; slow = slow->next; } quick = slow; slow = slow->next; quick->next = NULL; reverseList(slow);//将后半部分进行反转 quick = head; while(quick && slow)//将前半部分与反转后的后半部分进行合并 { ListNode *t = slow->next; slow->next = quick->next; quick->next = slow; slow = t; quick = quick->next->next; } } void reverseList(ListNode* &head)//采用头插法进行链表反转 { if(head == NULL || head->next == NULL) return ; ListNode *p = head->next; head->next = NULL; while(p) { ListNode *t = p->next; p->next = head; head = p; p = t; } } }; int main(void) { ListNode *head = new ListNode(6); addNode(head, 5); addNode(head, 4); addNode(head, 3); addNode(head, 2); addNode(head, 1); addNode(head, 0); printList(head); cout<<endl; Solution solution; solution.reorderList(head); printList(head); return 0; }
-----------------------我和我追猪的梦-----------------------------------------------------------------
作者:mickole
![](http://service.t.sina.com.cn/widget/qmd/1662683674/9b75a6f7/1.png)