143. Reorder List

• Total Accepted: 84681
• Total Submissions: 341465
• Difficulty: Medium
• Contributors: Admin

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}.

来源： https://leetcode.com/problems/reorder-list/

分析

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/**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */     /**    * slow fast pointer   * [1,2,3,4] 2   * [1,2,3,4,5] 3   * [1,2,3,4,5,6] 3   * ...   */ class Solution { public:     void reorderList(ListNode* head) {         if(head == NULL || head->next == NULL) return;                   ListNode* slow = head, * fast = head;         while(fast && fast->next && fast->next->next){             slow = slow->next;             fast = fast->next->next;         }                   ListNode* l1 = head;         ListNode* l2 = reverse(slow->next);         slow->next = NULL;                   ListNode dummy(0);         ListNode* p = &dummy;         bool sign = false;         while(l2){             if(sign){                 p->next = l2;                 l2 = l2->next;             }             else{                 p->next = l1;                 l1 = l1->next;             }             p = p->next;             sign = !sign;         }         p->next = l1;               }     ListNode* reverse(ListNode* head){         if(head == NULL || head->next == NULL) return head;         ListNode* p = NULL, * cur = head;         while(cur){             ListNode* tmp = cur->next;             cur->next = p;             p = cur;             cur = tmp;         }         return p;     } };

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