[LeetCode] Reorder List

最新推荐文章于 2019-10-17 10:50:20 发布

原创最新推荐文章于 2019-10-17 10:50:20 发布 · 273 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

131 篇文章

订阅专栏

本文介绍了一种链表重新排列算法，通过获取中间节点、反转后半部分链表并交错连接前半部分与反转后的后半部分，实现链表的特定顺序排列。此过程不修改节点值，直接在原有链表上操作。

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}.

class Solution {
public:
    void reorderList(ListNode *head) {
        if(head == NULL || head -> next == NULL) return;
        ListNode *mid = getMid(head);
        ListNode *back = reverse(mid -> next);
        mid -> next = NULL;
        ListNode *cur = new ListNode(-1);
        while(head && back){
            ListNode *tmpa = head -> next;
            head -> next = back;
            ListNode *tmpb= back ->next;
            back ->next = tmpa;
            head = tmpa;
            back = tmpb;
        }
    }
    ListNode *getMid(ListNode *head){
        ListNode *p = head;
        ListNode *q = head;
        while(q -> next && q -> next ->next){
            p = p -> next;
            q = q -> next ->next;
        }
        return p;
    }
    
    ListNode *reverse(ListNode *head){
        if(head == NULL)
            return NULL;  
          
        ListNode* p = head->next;  
        head->next = NULL;  
          
        while(p != NULL){  
            ListNode* tmp = p;  
            p = p->next;  
            tmp->next = head;  
            head = tmp;  
        }  
        return head;  
    }
};