leecode 978. Longest Turbulent Subarray(最长连续波动序列,DP or 滚动数组)

时间:2022-09-29 22:03:20

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978. Longest Turbulent Subarray

A subarray A[i], A[i+1], ..., A[j] of A is said to be turbulent if and only if:

  • For i <= k < jA[k] > A[k+1] when k is odd, and A[k] < A[k+1] when k is even;
  • OR, for i <= k < jA[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. 1 <= A.length <= 40000
  2. 0 <= 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。