标签：longest max lse class note else 子序列 ems tps
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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题意:求最长的连续波动子序列，注意是连续。
思路：DP滚动一下就行了。
class Solution { public: int maxTurbulenceSize(vector<int>& A) { int ans = 1; int dp[40001][2]; dp[0][1] = dp[0][0] = 1; for(int i = 1 ; i < A.size() ; i++){ if(A[i] > A[i-1]){ dp[i][0] = dp[i-1][1] + 1; dp[i][1] = 1; } else if(A[i] < A[i-1]){ dp[i][1] = dp[i-1][0] + 1; dp[i][0] = 1; } else{ dp[i][1] = 1; dp[i][0] = 1; } ans = max(ans,max(dp[i][0],dp[i][1])); } return ans; } };
那么，换个思路，如果求的是最长的波动序列呢（可不连续）？
改下DP就行了，看下面代码：
if(a[i]>a[i-1]){ dp[i][0]=max(dp[i-1][0],dp[i-1][1]+1); dp[i][1]=dp[i-1][1]; } else if(a[i]<a[i-1]){ dp[i][1]=max(dp[i-1][1],dp[i-1][0]+1); dp[i][0]=dp[i-1][0]; } else if(a[i]==a[i-1]){ dp[i][0]=dp[i-1][0]; dp[i][1]=dp[i-1][1]; }
return max(dp[n][0],dp[n][1]);
不是很难理解，递推下来不满足的不是等于1，而是等于上个状态取到的最长的。
以上。
leecode 978. Longest Turbulent Subarray(最长连续波动序列，DP or 滚动数组)
标签：longest max lse class note else 子序列 ems tps
原文地址：https://www.cnblogs.com/Esquecer/p/10311917.html
