package LeetCode_978 /** * 978. Longest Turbulent Subarray * https://leetcode.com/problems/longest-turbulent-subarray/description/ * * A subarray A[i], A[i+1], ..., A[j] of A is said to be turbulent if and only if: *For i <= k < j, A[k] > A[k+1] when k is odd, and A[k] < A[k+1] when k is even; *OR, for i <= k < j, A[k] > A[k+1] when k is even, and A[k] < A[k+1] when k is odd. That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray. Return the length of a maximum size turbulent subarray of A. Example 1: Input: [9,4,2,10,7,8,8,1,9] Output: 5 Explanation: (A[1] > A[2] < A[3] > A[4] < A[5]) * */ class Solution { /* * solution: Sliding Window, Time complexity:O(n), Space complexity:O(1) * */ fun maxTurbulenceSize(A: IntArray): Int { if (A == null || A.isEmpty()) { return 0 } if (A.size == 1) { return 1 } val len = A.size var max = 0 var left = 0 var right = 1 while (right < len) { if (isZigZag(A, left, right)) { max = Math.max(max, right - left + 1) right++ continue } //if the two number in left and right are equal //for example:[9,9] if (A[right] == A[right - 1]) { left = right max = Math.max(max, right - left + 1) right++ continue } left = right - 1 right++ } return max } //4,2,10,7,8 is zigzag private fun isZigZag(array: IntArray, start: Int, end: Int): Boolean { if (array[start] == array[start + 1]) { return false } var increasing = array[start] < array[start + 1] for (i in start + 1 until end) { //turn it each time increasing = !increasing if (increasing && array[i] >= array[i + 1] || !increasing && array[i] <= array[i + 1]) { return false } } return true } }