LeetCode: Swap Nodes in Pairs 解题报告

Swap Nodes in Pairs

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 be changed.

SOLUTION 1:

递归解法：

1. 先翻转后面的链表，得到新的Next.

2. 翻转当前的2个节点。

3. 返回新的头部。

// Solution 1: the recursion version.
    public ListNode swapPairs1(ListNode head) {
        if (head == null) {
            return null;
        }
        
        return rec(head);
    }
    
    public ListNode rec(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode next = head.next.next;
        
        // 翻转后面的链表
        next = rec(next);
        
        // store the new head.
        ListNode tmp = head.next;
        
        // reverse the two nodes.
        head.next = next;
        tmp.next = head;
        
        return tmp;
    }

SOLUTION 2:

迭代解法：

1. 使用Dummy node保存头节点前一个节点

2. 记录翻转区域的Pre（上一个节点），记录翻转区域的next，或是tail。

3. 翻转特定区域，并不断前移。

有2种写法，后面一种写法稍微简单一点，记录的是翻转区域的下一个节点。

// Solution 2: the iteration version.
    public ListNode swapPairs(ListNode head) {
        // 如果小于2个元素，不需要任何操作
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        // The node before the reverse area;
        ListNode pre = dummy;
        
        while (pre.next != null && pre.next.next != null) {
            // The last node of the reverse area;
            ListNode tail = pre.next.next;
            
            ListNode tmp = pre.next;
            pre.next = tail;
            
            ListNode next = tail.next;
            tail.next = tmp;
            tmp.next = next;
            
            // move forward the pre node.
            pre = tmp;
        }
        
        return dummy.next;
    }

// Solution 3: the iteration version.
    public ListNode swapPairs3(ListNode head) {
        // 如果小于2个元素，不需要任何操作
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        // The node before the reverse area;
        ListNode pre = dummy;
        
        while (pre.next != null && pre.next.next != null) {
            ListNode next = pre.next.next.next;
            
            ListNode tmp = pre.next;
            pre.next = pre.next.next;
            pre.next.next = tmp;
            
            tmp.next = next;
                        
            // move forward the pre node.
            pre = tmp;
        }
        
        return dummy.next;
    }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/list/SwapPairs3.java