LeetCode刷题笔记（链表）：reorder-list

转载请注明作者和出处：http://blog.csdn.net/u011475210

代码地址：https://github.com/WordZzzz/Note/tree/master/LeetCode

刷题平台：https://www.nowcoder.com/ta/leetcode

题  库：Leetcode经典编程题

编  者：WordZzzz

题目描述

解题思路

C版代码实现

题目描述

Given a singly linked list L: L 0→L 1→…→L n-1→L n,

reorder it to: L 0→L n →L 1→L n-1→L 2→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}.

解题思路

要求in-place，所以不能使用辅助空间。可以采用快慢指针先找到中间点，将原列表分成两个部分，然后再将后半部分反转链表，最后再根据题目规则一前一后合并成一个链表。

C++版代码实现

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

ListNode *getMiddle(ListNode *head){
ListNode *fast = head;
ListNode *slow = head;
while(fast->next != NULL && fast->next->next != NULL){
slow = slow->next;
fast = fast->next->next;
}
return slow;
}

ListNode *reverseList(ListNode *head){
ListNode *pReversedHead = NULL;
ListNode *pNode = head;
ListNode *pPrev = NULL;

while(pNode != NULL){
ListNode *pNext = pNode->next;
if(pNext == NULL)
pReversedHead = pNode;
pNode->next = pPrev;
pPrev = pNode;
pNode = pNext;
}
return pReversedHead;
}

void mergeList(ListNode *left, ListNode *right){
while(left != NULL && right != NULL){
ListNode *curLeft = left->next;
ListNode *curRight = right->next;
left->next = right;
right->next = curLeft;
left = curLeft;
right = curRight;
}
}

void reorderList(ListNode *head) {
if(head == NULL || head->next == NULL)
return ;

ListNode *middle = getMiddle(head);
ListNode *right = reverseList(middle->next);
middle->next = NULL;
ListNode *left = head;
mergeList(left, right);
}
};

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