876. Middle of the Linked List

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