LeetCode Reorder List

最新推荐文章于 2019-10-17 10:50:20 发布

coobj2008

最新推荐文章于 2019-10-17 10:50:20 发布

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本文介绍了一种链表重组算法，该算法将给定的单链表按特定模式重新排列，而不改变节点值。通过中间断开形成两个链表，反转后半部分链表并将其与前半部分交错合并。

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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}.

/**
 * 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 == NULL || head->next == NULL)
    		return;
    	ListNode *slow = head, *fast = head, *pre = head;
    	将列表从中间断开，形成两个链表head和slow 
		while(fast != NULL && fast->next != NULL) {
			fast = fast->next->next;
			pre = slow;
			slow = slow->next;	
		}
		pre->next = NULL;
		slow = reverseList(slow);
		mergeList(head, slow);		
    }
private:
	//将后部分链表反序 
	ListNode *reverseList(ListNode *head) {
		ListNode *cur = head;
		ListNode *pre = NULL;
		ListNode *post = cur->next;
		while(cur != NULL) {
			//记下下一个指针 
			post = cur->next;
			cur->next = pre;
			pre = cur;
			cur = post;
		}
		return pre;			
	}
	void mergeList(ListNode *head, ListNode *slow) {
		ListNode *tmp = head;
		ListNode *pre_tmp = head;
		ListNode *post_slow = slow->next;
		while(tmp != NULL) {
			post_slow = slow->next;
			slow->next = tmp->next; 
			tmp->next = slow;
			pre_tmp = tmp->next;
			tmp = tmp->next->next;
			slow = post_slow;
		} 
		if(slow != NULL)
			pre_tmp->next = slow; 
	}
};