【Leetcode -1721.交换链表中的节点 -2058.找出临界点之间的最小和最大距离】

		struct ListNode* swapNodes(struct ListNode* head, int k)
		{
		    //front为交换的两个节点的前一个节点，behind为交换的两个节点的后一个节点，cur用来让两个节点找到交换的两个位置
		    struct ListNode* front = head, * behind = head, * cur = head;
		
		    //front找到交换的第一个节点
		    while (--k)
		    {
		        front = front->next;
		        cur = cur->next;
		    }
		
		    //behind找到交换的第二个节点
		    while (cur->next)
		    {
		        cur = cur->next;
		        behind = behind->next;
		    }
		
		    //找到两个交换的节点之后，就对它们进行换值
		    int num = front->val;
		    front->val = behind->val;
		    behind->val = num;
		
		    return head;
		}

		int* nodesBetweenCriticalPoints(struct ListNode* head, int* returnSize)
		{
		    struct ListNode* cur = head;
		
		    //minDistance为最小距离，maxDistance为最大距离，index记录下标
		    int minDistance = -1, maxDistance = -1, index = 1;
		
		    //返回两个整型
		    int* ret = (int*)malloc(sizeof(int) * 2);
		    *returnSize = 2;
		
		    //数组存放临界点的下标，len记录当前数组的长度
		    int arr[100000] = { 0 };
		    int len = 0;
		
		    //每次比较前一个的val和后一个val，若符合题意就放入数组
		    while (cur->next && cur->next->next)
		    {
		        int x = cur->val;
		        cur = cur->next;
		        index++;
		        if (cur->val > x && cur->val > cur->next->val)
		            arr[len++] = index;
		
		        else if (cur->val < x&& cur->val < cur->next->val)
		            arr[len++] = index;
		    }
		
		    //len，数组长度大于1，据题意，即临界点要大于两个
		    if (len > 1)
		    {
		        //上面len多加了一次，所以要减一次
		        len--;
		
		        //maxDistance即为最大减最小
		        maxDistance = abs(arr[len] - arr[0]);
		
		        //minDistance要遍历一次数组，找出相邻之间最小的值
		        for (int i = 1; i <= len; i++)
		        {
		            if (i == 1)
		                minDistance = arr[i] - arr[i - 1];
		            minDistance = minDistance > arr[i] - arr[i - 1] ? arr[i] - arr[i - 1] : minDistance;
		        }
		    }
		
		
		    ret[0] = minDistance;
		    ret[1] = maxDistance;
		
		    return ret;
		}