题目链接：力扣

思路：用三个指针标记三个位置，从前往后找满足条件的小标，放在数组里面，最大距离好找，就是数组中最后一个和第一个的差，最小距离就是数组中相邻的两个值的差的最小值（最下距离，在循环的时候判断一下即可，这是一个坑点）~~

上代码：

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun nodesBetweenCriticalPoints(head: ListNode?): IntArray {
        val arrayIndex = ArrayList<Int>()
        var front = head
        var behind = head?.next
        var index = 1
        var minn = 1000000
        while (behind?.next != null) {
            val p = behind.next
            if (front?.`val`!! < behind.`val` && behind.`val` > p?.`val`!!
                    || front.`val` > behind.`val` && behind.`val` < p?.`val`!!) {
                arrayIndex.add(index)
                if (arrayIndex.size > 1) {
                    minn = minn.coerceAtMost(arrayIndex[arrayIndex.size - 1] - arrayIndex[arrayIndex.size - 2])
                }
            }
            front = behind
            behind = p
            index++
        }
        if (arrayIndex.size > 1) {
            return intArrayOf(minn, arrayIndex[arrayIndex.size - 1] - arrayIndex[0])
        }
        return intArrayOf(-1, -1)
    }
}