2014年微軟校園招聘筆試題最後編程題Reorder List

1.問題

Given a singly linked list L: (L0 , L1 , L2...Ln-1 , Ln). Write a program to reorder it so that it becomes(L0 , Ln , L1 , Ln-1 , L2 , Ln-2...).
struct Node  
{  
    int val_;  
    Node* next;  
};  
Notes:
1）Space Complexity should be O(1) 
2）Only the ".next" field of a node is modifiable.
該題目和LeetCode中的Reorder List 題目完全一樣：
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}.

2.分析

如果採用空間複雜度爲O(n)和時間複雜度爲O(1)的方法解決Reorder List問題，主要分爲以下三步：
1）將單鏈表分爲前後兩個鏈表；
2）逆序分開後的第二個鏈表；
3）再將前後鏈表合併成一個單鏈表。

3.程序實現

/*
 *@題目描述： 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}.
*@問題分析：1.將鏈表分成兩部分；
*			2.將第二個鏈表逆轉；
*			3.將兩個鏈表合併。
 */
//鏈表的數據結構
class ListNode{
	ListNode next;
	int val;
	public ListNode(int val) {
		// TODO Auto-generated constructor stub
		this.val=val;
		next=null;
	}
}

public class ReorderList {
	public static ListNode reverseList(ListNode head){
		if(head==null || head.next==null)
			return head;
		ListNode preNode=head;
		ListNode currNode=head.next;
		while(currNode!=null){
			ListNode tempNode=currNode.next;
			currNode.next=preNode;
			preNode=currNode;
			currNode=tempNode;
		}
		head.next=null;
		return preNode;
	}
	public static void  reorderList(ListNode head){
		if(head!=null && head.next!=null){
			//將兩個鏈表從中間分爲兩個鏈表
			ListNode slowNode=head;
			ListNode fastNode=head;
			while(fastNode!=null&&fastNode.next!=null&&fastNode.next.next!=null){
				slowNode=slowNode.next;
				fastNode=fastNode.next.next;
			}
			ListNode secondListHead=slowNode.next;
			slowNode.next=null;
			//逆轉第二個鏈表
			secondListHead=reverseList(secondListHead);
			ListNode pHead1=head;
			ListNode pHead2=secondListHead;
			//合併兩個分開的鏈表
			while(pHead2!=null){
				ListNode pTempNode1=pHead1.next;
				ListNode pTempNode2=pHead2.next;
				pHead1.next=pHead2;
				pHead2.next=pTempNode1;
				pHead1=pTempNode1;
				pHead2=pTempNode2;
			}
		}
	}
	public static void printListInfo(ListNode head){
		System.out.println("*****開始分隔符*****");
		while(head!=null){
			System.out.print(head.val);
			head=head.next;
		}
		System.out.println("\n*****結束分隔符*****");
	}
	public static void main(String args[]){
		ListNode listHead=new ListNode(3);
		ListNode listNode1=new ListNode(4);
		ListNode listNode2=new ListNode(5);
		ListNode listNode3=new ListNode(6);
		ListNode listNode4=new ListNode(7);
		listHead.next=listNode1;
		listNode1.next=listNode2;
		listNode2.next=listNode3;
		listNode3.next=listNode4;
		System.out.println("Before Reorder list: ");
		printListInfo(listHead);
		reorderList(listHead);
		System.out.println("after Reorder List:");
		printListInfo(listHead);
	}
}

運行結果如下：

Before Reorder list: 
*****開始分隔符*****
34567
*****結束分隔符*****
after Reorder List:
*****開始分隔符*****
37465
*****結束分隔符*****