lintcode-easy-Delete Node in the Middle of Singly Linked List

Implement an algorithm to delete a node in the middle of a singly linked list, given only access to that node.

Example

Given 1->2->3->4, and node 3. return 1->2->4

删除的方法就是用后面的值覆盖前面的值，注意避免OBOB

/*** Definition for ListNode.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int val) {*         this.val = val;*         this.next = null;*     }* }*/
public class Solution {/*** @param node: the node in the list should be deleted* @return: nothing*/public void deleteNode(ListNode node) {// write your code hereListNode p = node;while(p.next.next != null){p.val = p.next.val;p = p.next;}p.val = p.next.val;p.next = null;return;}
}

转载于:https://www.cnblogs.com/goblinengineer/p/5213808.html

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