[LeetCode]Reorder List

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}.

Analysis:

One straightforward middle step of such reordering is:
{1,2,3,4,5,6}  --> {1,2,3,6,5,4} --> {1,6,2,5,3,4}
{1,2,3,4,5,6,7}---> {1,2,3,4,7,6,5} ---> {1,7,2,6,3,5,4}

By reversing the last half part of the linked list, we do not need to worried about the "parent" pointer anymore. The final step is just insert the each element in the last half part into the first part (every two element).

So the algorithm implemented below can be summarized as:
Step 1  Find the middle pointer of the linked list (you can use the slow/fast pointers)
Step 2  Reverse the second part of the linked list (from middle->next to the end)
Step 3  Do the reordering. (inset every element in the second part in between the elements in the first part)

java

public void reorderList(ListNode head) {        if(head==null || head.next == null|| head.next.next == null) return;        ListNode slow = head;        ListNode fast = head;        while(fast!=null&& fast.next!=null){        slow = slow.next;        fast = fast.next.next;        }        //slow is in the middle, reverse slow.next to end        fast = slow.next;        slow.next = null;        while(fast!=null){        ListNode temp = fast.next;        fast.next = slow;        slow = fast;        fast = temp;        }//merge two list        ListNode first = head;        ListNode second = slow;        while(first!=null && first.next!=second && first!=second){        ListNode temp = second.next;        second.next = first.next;        first.next = second;        first = second.next;        second = temp;        }    }

c++

void reorderList(ListNode *head) {        if(head == NULL || head->next == NULL) return;        //find half node        ListNode *slow = head;        ListNode *fast = head;        while(true){            fast = fast->next;            if(fast == NULL)                break;            fast = fast->next;            if(fast == NULL)                break;            slow = slow->next;        }        //resever post half list        ListNode *cur = slow;        ListNode *pre = slow->next;        cur->next = NULL;        while(pre!=NULL){            ListNode *temp = pre->next;            pre->next = cur;            cur = pre;            pre = temp;        }        // merge two list        ListNode *first = head;        ListNode *second = cur;        while(first != second && first != NULL && second != NULL){            ListNode *temp = second->next;            second->next = first->next;            first->next = second;            first = second->next;            second = temp;        }            }