标签：last one output only ges problems reduce some max
原题链接在这里：https://leetcode.com/problems/longest-turbulent-subarray/
题目：
A subarray A[i], A[i+1], ..., A[j] of A is said to be turbulent if and only if:
i <= k < j, A[k] > A[k+1] when k is odd, and A[k] < A[k+1] when k is even;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])
Example 2:
Input: [4,8,12,16]
Output: 2
Example 3:
Input: [100]
Output: 1
Note:
1 <= A.length <= 400000 <= A[i] <= 10^9题解：
Set some small examples like [1, 3, 2], [2,2] and find routine.
It matters the last 3 componenets. If it is l<m>r or l>m<r relationship, then length+1. Otherwise, reset to 2 or 1.
Let dp[i] denotes up to A[i-1], the longest turbulent length.
If A[i-3]<A[i-2]>A[i-1] or A[i-3]>A[i-2]<A[i-1], dp[i] = dp[i-1] + 1.
Maintain the maximum to res.
Time Complexity: O(n). n = A.length.
Space: O(n).
AC Java:
1 class Solution { 2 public int maxTurbulenceSize(int[] A) { 3 if(A == null){ 4 return 0; 5 } 6 7 if(A.length < 2){ 8 return A.length; 9 } 10 11 int len = A.length; 12 int [] dp = new int[len+1]; 13 dp[1] = 1; 14 dp[2] = A[0] == A[1] ? 1 : 2; 15 16 int res = dp[2]; 17 for(int i = 3; i<=len; i++){ 18 if(A[i-2]<A[i-3] && A[i-2]<A[i-1] || A[i-2]>A[i-3] && A[i-2]>A[i-1]){ 19 dp[i] = dp[i-1] + 1; 20 res = Math.max(res, dp[i]); 21 }else if(A[i-1] == A[i-2]){ 22 dp[i] = 1; 23 }else{ 24 dp[i] = 2; 25 } 26 } 27 28 return res; 29 } 30 }
It only cares about dp[i-1]. Thus it could reduce dimension.
Time Complexity: O(n).
Space: O(1).
AC Java:
1 class Solution { 2 public int maxTurbulenceSize(int[] A) { 3 if(A == null){ 4 return 0; 5 } 6 7 if(A.length < 2){ 8 return A.length; 9 } 10 11 int len = A.length; 12 int dp = A[0] == A[1] ? 1 : 2; 13 int res = dp; 14 15 for(int i = 3; i<=len; i++){ 16 if(A[i-2]<A[i-3] && A[i-2]<A[i-1] || A[i-2]>A[i-3] && A[i-2]>A[i-1]){ 17 dp = dp + 1; 18 res = Math.max(res, dp); 19 }else if(A[i-1] == A[i-2]){ 20 dp = 1; 21 }else{ 22 dp = 2; 23 } 24 } 25 26 return res; 27 } 28 }
LeetCode 978. Longest Turbulent Subarray
标签：last one output only ges problems reduce some max
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11484108.html
