[LeetCode]Swap Nodes in Pairs

版权声明：本文为博主原创文章。未经博主同意不得转载。 https://blog.csdn.net/yeweiouyang/article/details/37532101

题目：给定一个单链表，要求对链表每两两节点进行交换，样例： 1->2->3->4, 交换后 2->1->4->3.

算法：链表操作，设置4个指针

pre：左边界，维护交换后链表的结构

swapA：指向要交换的第一个节点

swapB：指向要交换的第二个节点

post：右边界，维护交换后链表的结构

链表：...->1->2->3->4

相应：pre  swapA  swapB  post

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
	        if (null == head) {
	        	return null;
	        }
	    	
	    	ListNode pre = head;
	        ListNode post = null;
	        ListNode swapA = head;
	        ListNode swapB = null;
	        while (null != pre) {
	        	if (head == pre) {
	        		// begin with the list head
	        		if (null != pre.next) {
	        			swapA = pre;
	        			swapB = swapA.next;
	        			post = swapB.next;
	        			
	        			// swap node pair
	        			swapA.next = post;
	        			swapB.next = swapA;
	        			head = swapB;
	        		} else {
	        			break;
	        		}
	        	} else {
//	        		pre = swapA;  // after swap
	        		if (null != pre.next) {
		        		swapA = pre.next;
		        		if (null != swapA.next) {
		        			swapB = swapA.next;
		        			post = swapB.next;
		        			
		        			// swap node pair
		        			pre.next = swapB;
		            		swapB.next = swapA;
		            		swapA.next = post;
		        		} else {
		        			break;  // last one node, don't swap
		        		}
		        	} else {
		        		break;  // end of the list
		        	}
	        	}
	        	
	        	pre = swapA;  // after swap
	        }
	        
	        return head;
	    }
}