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Given a linked list and two values v1 and v2. Swap the two nodes in the linked list with values v1 and v2. It‘s guaranteed there is no duplicate values in the linked list. If v1 or v2 does not exist in the given linked list, do nothing.
You should swap the two nodes with values v1 and v2. Do not directly swap the values of the two nodes.
Given 1->2->3->4->null
and v1 = 2
, v2 = 4
.
Return 1->4->3->2->null
.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { /** * @param head a ListNode * @oaram v1 an integer * @param v2 an integer * @return a new head of singly-linked list */ public ListNode swapNodes(ListNode head, int v1, int v2) { // Write your code here if(head == null || head.next == null) return head; if(v1 == v2) return head; ListNode fakehead = new ListNode(0); fakehead.next = head; ListNode p1 = fakehead; ListNode p2 = fakehead; while(p1.next != null){ if(p1.next.val == v1) break; p1 = p1.next; } if(p1.next == null) return head; while(p2.next != null){ if(p2.next.val == v2) break; p2 = p2.next; } if(p2.next == null) return head; ListNode node1 = p1.next; p1.next = null; ListNode node2 = p2.next; p2.next = null; p2.next = node1; p1.next = node2; ListNode head2 = null; ListNode head3 = null; if(node1.next != null) head2 = node1.next; else head2 = null; if(node2.next != null) head3 = node2.next; else head3 = null; node2.next = head2; node1.next = head3; return fakehead.next; } }
lintcode-medium-Swap Two Nodes in Linked List
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原文地址:http://www.cnblogs.com/goblinengineer/p/5362034.html