leetcode: Reorder List

最新推荐文章于 2018-04-22 21:11:48 发布

原创最新推荐文章于 2018-04-22 21:11:48 发布 · 340 阅读

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本文介绍了一种链表重组算法，该算法将一个单链表L0→L1→…→Ln-1→Ln重新排列为L0→Ln→L1→Ln-1→L2→Ln-2→…。通过在不改变节点值的情况下实现链表的重新排序，并提供了具体的实现代码示例。

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Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→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}.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {

public:
    ListNode* reverse(ListNode* node) {
        
        if (!node || !node->next) {
            return node;
        }
        
        ListNode *pPre, *pCur, *pNext;
        
        pPre = node;
        pCur = pPre->next;
        pPre->next = NULL;
        
        while (pCur && pCur->next) {
            pNext = pCur->next;
            
            pCur->next = pPre;
            pPre = pCur;
            pCur = pNext;
        }
        
        pCur->next = pPre;
        
        return pCur;
    }
    


    void reorderList(ListNode *head) {
        
        ListNode *slow;
        ListNode *fast;
        ListNode *temp;
        
        if (!head || !head->next)
            return;
        
        slow = fast = head;
        
        while (fast && fast->next)
        {
            temp = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        
        temp->next = NULL;
        
        ListNode *Half = slow;
        ListNode *newHalf = reverse(Half);
        
        
        ListNode *p1 = head;
        ListNode *p2 = newHalf;
        ListNode *t1;
        ListNode *t2;
        ListNode *p = p1;
        
        while (p1 && p2)
        {
            t1 = p1->next;
            p1->next=p2;
            p = p1->next;
            p1 = t1;
            
            
            t2 = p2->next;
            if (p1)
            {
                p2->next=p1;
            }
            p = p2->next;
            p2 = t2;
            
        }
        
        if (p1)
            p->next=p1;
    }
};