24. 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.

是 http://www.jianshu.com/p/6b64490cac79 的简单版本when k =2

Solution1：Recursive: 先序处理

思路：发现pattern子问题，处理好当前的，递归剩下的
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

屏幕快照 2017-09-11 下午12.51.38.png

Solution2：Iterative

思路：遍历每次处理一个section(两个nodes)
Time Complexity: O(N) Space Complexity: O(1)

屏幕快照 2017-09-11 下午1.07.26.png

Solution2.2 Round1：Iterative

Solution1 Code：

class Solution1 {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null)
            return head;
        
        ListNode second = head.next;
        
        head.next = swapPairs(second.next);
        second.next = head;
        return second;
    }
}

Solution2 Code：

class Solution2 {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode current = dummy;
        while (current.next != null && current.next.next != null) {
            ListNode first = current.next;
            ListNode second = current.next.next;
            
            first.next = second.next;
            second.next = first;
            current.next = second;
            
            current = current.next.next;
        }
        return dummy.next;
    }
}

Solution2.2 Round1 Code：

class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode cur = dummy;
        
        while(cur != null) {
            cur = swapPair(cur);
        }
        
        return dummy.next;
    }
    
    private ListNode swapPair(ListNode pre) {
        if(pre.next == null || pre.next.next == null) {
            return null;
        }
        ListNode cur = pre.next;
        ListNode post = cur.next;
        ListNode save = post.next;
        
        pre.next = post;
        post.next = cur;
        cur.next = save;
        
        return cur;
    }
}