Problem:
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.
思路：
Solution (C++):
ListNode* middleNode(ListNode* head) {
int len = 0;
ListNode* cur = head;
while (cur) {
++len;
cur = cur->next;
}
int n = len / 2;
cur = head;
while (n) {
cur = cur->next;
--n;
}
return cur;
}
性能：
Runtime: 0 ms Memory Usage: 6.4 MB
思路：
Solution (C++):
int peakIndexInMountainArray(vector<int>& A) {
for (int i = 0; i < A.size()-1; ++i) {
if (A[i] > A[i+1]) return i;
}
return -1;
}
性能：
Runtime: 12 ms Memory Usage: 7.4 MB
