LeetCode(143)Reorder List

题目如下：

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}.

分析如下：

写几个例子，发现可以这么来实现题目要求的转换。首先，把链表分成前半部分和后半部分，然后把后半部分反序，最后把后半部分依次插入前半部分。

我的代码：

//280ms过大集合
/**
 * 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||head->next->next==NULL)
            return;
        //find the first half( may be either odd or even numbers)
        int n=0;
        ListNode* p=head;
        while(p!=NULL){
            n=n+1;
            p=p->next;
        }
        
        int half_point=n/2+n%2;
        p=head;
        n=1;
        while(n!=half_point){
            p=p->next;
            n++;
        }
        //reverse the second half
        ListNode* second_half_old_start=p->next;
        p->next=NULL;//bug1 断开list要在断点出赋NULL
        p=second_half_old_start;
        ListNode* q=second_half_old_start->next;
        ListNode* r=q;
        while(q!=NULL){
            r=q->next;
            q->next=p;
            p=q;
            q=r;
        }
        second_half_old_start->next=NULL;
        // insert the second half into the first half.
        q=head;
        while(p!=NULL){
            r=p->next; //bug2 用r保持当前的链表，否则丢链
            p->next=q->next;
            q->next=p;
            p=r;
            q=q->next->next;
        }
        return;
    }
};

小结:

(1) 非IDE环境下手写，还是有若干bug，哎。

update: 2015-01-05 思路和上面是一样的。

/*
 *找到2分点。左边一半保留，右边一半反转，新的右边链表逐个加入到左边链表中。
 */
//63ms
class Solution {
public:
    void reorderList(ListNode *head) {
        if (head == NULL || head->next == NULL || head->next->next == NULL ) return; //should be at least 3 nodes
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* first_half_head = head;
        ListNode* first_half_tail = NULL;
        ListNode* second_half_head = NULL;
        while (fast != NULL && fast->next != NULL) {
            fast = fast->next->next;
            slow = slow->next;
        }
        
        first_half_tail = slow;
        second_half_head = slow->next;
        first_half_tail->next = NULL;
        ListNode* p = second_half_head;
        ListNode* q = second_half_head->next;
        ListNode* r = NULL;
        while (q != NULL) {
            r = q->next;
            q->next = p;
            p = q;
            q = r;
        }
        second_half_head->next = NULL; // new tail
        second_half_head = p; //new head;
        ListNode* t = head;
        ListNode* m = second_half_head;
        while(second_half_head != NULL) {
            m = m->next;
            second_half_head->next = head->next;
            head->next = second_half_head;
            head = head->next->next;
            second_half_head = m;
        }
        head = first_half_head;
    }
};