lintcode 451

https://www.lintcode.com/problem/451/

[cost]
30 min

[brainstorm]
1 decouple corner case and general case handling.
2 when moving to next two nodes, what case should be considered?
3 how to keep head as return value?

[desc]
1 handle corner cases.
2 handle general case.
keep second node as head to return.
point 1st node from second node.
point 1st node to third fourth nodes block.
go to handle next two nodes.

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: a ListNode
     * @return: a ListNode
     */
    public ListNode swapPairs(ListNode head) {
        // write your code here
        if(head==null || head.next==null){
            return head;
        }
        ListNode fir = head;
        ListNode sec = fir.next;
        ListNode ret = sec;
        while(fir!=null && sec!=null){
            ListNode third = sec.next;
            //swap
            sec.next = fir;
            fir.next = third;
            ListNode last = fir;

            //next two nodes
            fir = third;
            sec = fir==null?null:fir.next;
            last.next = sec==null?fir:sec;              
        }

        return ret;
            
    }
}

标签：fir,head,ListNode,451,lintcode,next,sec,null
来源： https://blog.csdn.net/laohixdxm00/article/details/120222036