143. Reorder List

不定期更新leetcode解题java答案。

采用pick one的方式选择题目。

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

如题意，按照如上方式将单链表重新排序。由于单链表的特性，如果用正常方式逐个改变链表位置，需要做多次遍历获取节点信息。因此，首先将后半链表倒序处理，再将分开的两条链表进行链接。

代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        //count nodes    
        ListNode node = head;
        int count = 0;
        while(node != null){
            count++;
            node = node.next;
        }
        if(count <= 1)
            return;
        
        //get middle node
        ListNode last = null;
        ListNode now = head;
        for(int i = 0; i < count / 2; i++){
            if(i == count / 2 - 1){
                ListNode tmp = now.next;
                now.next = null;
                now = tmp;
            }else
                now = now.next;
        }
        ListNode next = now.next;
        //reverse the nodes after the mid node
        while(next != null){
            ListNode tmp = next.next;
            now.next = last;
            last = now;
            now = next;
            next = tmp;
        }
        now.next = last;    //need point one more time
        
        ListNode mid = now; //the new middle node after reverse
        
        node = head;
        //combine two linked lists
        while(node != null){
            ListNode tmp = node.next;
            
            node.next = mid;
            node = tmp;
            
            tmp = mid.next;
            if(node != null)
                mid.next = node;
            mid = tmp;
        }
    }
}

由于本题要求不允许直接改变链表数值，不应采用此方法，亦将代码贴上，测试可得下述代码时间消耗更长：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
 public class Solution {
    public void reorderList(ListNode head) {
        ListNode node = head;
        ArrayList<Integer> list = new ArrayList();
        while(node != null){
            list.add(node.val);
            node = node.next;
        }
        
        for(int i = 0; i < list.size(); i++){
            head.val = i % 2 == 0 ? list.get(i / 2) : list.get(list.size() - 1 - i / 2);
            head = head.next;
        }
    }
}