143. Reorder List

Given a singly linked list L: L0 → L1 → … → Ln-1 → Ln

reorder it to: L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …

Example

Given 1->2->3->4->null, reorder it to 1->4->2->3->null.

分析：

先得到后一半，然后reverse后一半，再和前一半合并。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: void
     */
    public void reorderList(ListNode head) {  
        if (head == null || head.next == null) return;
        
        ListNode slow = head;
        ListNode quick = head;
        
        while (quick.next != null && quick.next.next != null) {
            slow = slow.next;
            quick = quick.next.next;
        }
        
        ListNode secondHead = slow.next;
        slow.next = null;
        secondHead = reverse(secondHead);
        
        ListNode firstHead = head;
        
        while (secondHead != null) {
            ListNode next1 = firstHead.next;
            ListNode next2 = secondHead.next;
            firstHead.next = secondHead;
            secondHead.next = next1;
            firstHead = next1;
            secondHead = next2;
        }
    }
    
    public ListNode reverse(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode pre = head;
        ListNode cur = head.next;
        pre.next = null;
        
        while (cur != null) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

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