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  • LeetCode--Remove Linked List Element

    Remove all elements from a linked list of integers that have value val.

    Example
    Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
    Return: 1 --> 2 --> 3 --> 4 --> 5

    Solution1: 不使用头结点。  

    注意事项:while停止的条件要特备注意;各种输入情况都要考虑到;如果首节点就是要删除的节点肿么办,等等问题。

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ListNode removeElements(ListNode head, int val) {
             if(head == null)
                  return null;
              if(head.next==null && head.val==val)
                  return null;
              if(head.next==null && head.val!=val)
                  return head;
              ListNode p = head;
              
              while(head.next!=null && head.val==val){
                  p = head;
                  head = head.next;
                  p.next = null;
              }
              
              p = head;
              ListNode q = p.next;
              while(p.next!=null){
                  q = p.next;
                  if(q.val == val){
                      p.next = q.next;
                      q.next = null;
                  }
                  else{
                      p = q;
                      q = p.next;
                  }
              }
              if(head.val==val)//做最后的检查,看是否首节点是要删除的元素
                  return null;
              return head;
             
        }
    }

    Solution2: 使用附加头结点。

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ListNode removeElements(ListNode head, int val) {
           if(head == null || head.val == val && head.next == null) return null;
            ListNode h = new ListNode(-1);
            h.next = head;
            ListNode p = h;
            ListNode q;
            while(p.next != null) {
                if(p.next.val == val) {
                    q = p.next;
                    p.next = q.next;
                    q.next = null;
                }
                else 
                    p = p.next;
            }
            return h.next;
        }
    }
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  • 原文地址:https://www.cnblogs.com/little-YTMM/p/4511473.html
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