Leetcode-Swap Nodes in Pairs

作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/23971481
时间：2014-4-17

题目

Swap Nodes in Pairs

 Total Accepted: 11114 Total Submissions: 35009My Submissions

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can

 Definition for singly-linked list. public class ListNode {     int val;     ListNode next;     ListNode(int x) {         val = x;         next = null;     } } 

解法

一个错误的想法

奇数偶数解法(可以但是写错)

 private ListNode swapPairs1(ListNode head){        if(head==null||head.next == null)        return head;         ListNode retNode =new ListNode(0);        retNode.next = head;        ListNode oddNode = head;        ListNode evenNode = head.next;        int count =0;        ListNode node =retNode;               while(oddNode!=null && evenNode!=null){             ListNode nexth = evenNode.next;                        node.next=evenNode;            evenNode.next=oddNode;            node = node.next.next;                                    if(nexth!=null&&nexth.next==null){// num of list is odd                node.next=nexth;                break;            }            if(nexth==null){                break;            }            oddNode = nexth;            evenNode = nexth.next;        }                      return retNode.next;    }

精妙解法

public class Solution {    public ListNode swapPairs(ListNode head) {        return swapPairs2(head);    }    private ListNode swapPairs2(ListNode head){        ListNode dummy = new ListNode(0);        dummy.next = head;        ListNode p = dummy;        //(0-p) -(1-h)-2-3-4-5        while(p.next != null && p.next.next != null){            ListNode q = p.next.next; //(0-p) -(1-pn)-(2-q)-3-4-5            p.next.next = q.next;//(0-p) -(1-pn)-(2-q)-(3-pnn)-4-5            q.next = p.next;//(0-p) -(1-pn-qn)-(2-q)-(3-pnn)-(4-qn)-5            p.next = q;//(0-p) -(1-qn)-(2-pn)-(3-pnn)-(4-qn)-5            p = p.next.next;//(0) -(1-qn)-(2-pn)-(3-p)-(4-qn)-5        }        return dummy.next;    }}

这么看其实还是不太懂，但是如果把测试用例打印出来就不难发现，其实p是用来调配结果的指针，q是用来调换位置的。

这道题需要用一定的空间来解，这题的空间效率大约是O（logn），非常精妙。

用q先存奇数位，p存偶数位，通过互相调换来移动。

测试用例大家也不必写了，做如下分享。

private ListNode swapPairs2(ListNode head){        ListNode dummy = new ListNode(-1);        dummy.next = head;        ListNode p = dummy;        //(0-p) -(1-h)-2-3-4-5        int whilecount = 1;        int linecount = 0;        while(p.next != null && p.next.next != null){        linecount =0;            ListNode q = p.next.next; //(0-p) -(1-pn)-(2-q)-3-4-5            printcode(++linecount,whilecount,"ListNode q = p.next.next;",p,q);//test                                              p.next.next = q.next;//(0-p) -(1-pn)-(2-q)-(3-pnn)-4-5            printcode(++linecount,whilecount,"p.next.next = q.next;",p,q);//test                                   q.next = p.next;//(0-p) -(1-pn-qn)-(2-q)-(3-pnn)-(4-qn)-5            printcode(++linecount,whilecount,"q.next = p.next;",p,q);//test                                    p.next = q;//(0-p) -(1-qn)-(2-pn)-(3-pnn)-(4-qn)-5            printcode(++linecount,whilecount,"p.next = q;",p,q);//test                                    p = p.next.next;//(0) -(1-qn)-(2-pn)-(3-p)-(4-qn)-5            printcode(++linecount,whilecount,"p = p.next.next;",p,q);//test                        System.out.println("------------------------------------------------------");            whilecount++;        }        return dummy.next;    }

private void printcode(int linecount,int count,String string,ListNode p, ListNode q) {// TODO Auto-generated method stubSystem.out.println("Line "+linecount+" Count"+count+":"+"code=="+string);        System.out.println("q:"+q.val);        testList.write(q);        System.out.println("p:"+p.val);        testList.write(p);}

public class testList {protected static void write(ListNode head){for(;head.next!=null;head=head.next)System.out.print(head.val+"->");System.out.println(head.val);}}

测试结果：

0->1->2->3->4->5->6->7->8
Line 1 Count1:code==ListNode q = p.next.next;
q:1
1->2->3->4->5->6->7->8
p:-1
-1->0->1->2->3->4->5->6->7->8
Line 2 Count1:code==p.next.next = q.next
q:1
1->2->3->4->5->6->7->8
p:-1
-1->0->2->3->4->5->6->7->8
Line 3 Count1:code==q.next = p.next;
q:1
1->0->2->3->4->5->6->7->8
p:-1
-1->0->2->3->4->5->6->7->8
Line 4 Count1:code==p.next = q;
q:1
1->0->2->3->4->5->6->7->8
p:-1
-1->1->0->2->3->4->5->6->7->8
Line 5 Count1:code==p = p.next.next;
q:1
1->0->2->3->4->5->6->7->8
p:0
0->2->3->4->5->6->7->8
------------------------------------------------------
Line 1 Count2:code==ListNode q = p.next.next;
q:3
3->4->5->6->7->8
p:0
0->2->3->4->5->6->7->8
Line 2 Count2:code==p.next.next = q.next
q:3
3->4->5->6->7->8
p:0
0->2->4->5->6->7->8
Line 3 Count2:code==q.next = p.next;
q:3
3->2->4->5->6->7->8
p:0
0->2->4->5->6->7->8
Line 4 Count2:code==p.next = q;
q:3
3->2->4->5->6->7->8
p:0
0->3->2->4->5->6->7->8
Line 5 Count2:code==p = p.next.next;
q:3
3->2->4->5->6->7->8
p:2
2->4->5->6->7->8
------------------------------------------------------
Line 1 Count3:code==ListNode q = p.next.next;
q:5
5->6->7->8
p:2
2->4->5->6->7->8
Line 2 Count3:code==p.next.next = q.next
q:5
5->6->7->8
p:2
2->4->6->7->8
Line 3 Count3:code==q.next = p.next;
q:5
5->4->6->7->8
p:2
2->4->6->7->8
Line 4 Count3:code==p.next = q;
q:5
5->4->6->7->8
p:2
2->5->4->6->7->8
Line 5 Count3:code==p = p.next.next;
q:5
5->4->6->7->8
p:4
4->6->7->8
------------------------------------------------------
Line 1 Count4:code==ListNode q = p.next.next;
q:7
7->8
p:4
4->6->7->8
Line 2 Count4:code==p.next.next = q.next
q:7
7->8
p:4
4->6->8
Line 3 Count4:code==q.next = p.next;
q:7
7->6->8
p:4
4->6->8
Line 4 Count4:code==p.next = q;
q:7
7->6->8
p:4
4->7->6->8
Line 5 Count4:code==p = p.next.next;
q:7
7->6->8
p:6
6->8

------------------------------------------------------

返回

LeetCode Solution(持续更新，java>c++)