2058. Find the Minimum and Maximum Number of Nodes Between Critical Points

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    vector<int> nodesBetweenCriticalPoints(ListNode* head)
    {
        vector<int> res = {-1, -1};
        int minDist = INT_MAX;

        ListNode* prev = head;
        ListNode* cur = head->next;

        int i = 1;
        int last = 0;
        int first = 0;

        while (cur->next)
        {
            // if (localMinimum || localMaximum)
            if ((cur->val < prev->val && cur->val < cur->next->val) || (cur->val > prev->val && cur->val > cur->next->val))
            {
                // 如果這是第一個 critial point，則將 last 和 first 都設為當前的 index i。
                if (last == 0)
                {
                    last = i;
                    first = i;
                }
                else
                {
                    minDist = min(minDist, i - last);
                    // 更新 last critial point
                    last = i;
                }
            }
            i++;
            // update the pointer
            prev = cur;
            cur = cur->next;
        }

        if (minDist != INT_MAX)
        {
            // maxDist 即為 last - first
            int maxDist = last - first;
            res = {minDist, maxDist};
        }
        return res;
    }
};

• T: $O(N)$
• S: $O(1)$