题目描述
Given a non-empty, singly linked list with head node head, return a middle node of linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.
Note:
The number of nodes in the given list will be between 1 and 100.
我的解法
解题思路
使用双指针，慢指针一次走一步，快指针一次走两步，当快指针走到了头的时候，慢指针就到了链表的中间。1
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* middleNode(ListNode* head) {
ListNode* p1 = head;
ListNode* p2 = head;
while(p2 && p2 -> next){
p1 = p1 -> next;
p2 = p2 -> next -> next;
}
return p1;
}
};
执行用时 :0 ms, 在所有 C++ 提交中击败了100.00%的用户
内存消耗 :8 MB, 在所有 C++ 提交中击败了100.00%的用户