[leetcode-376-Wiggle Subsequence]

Posted 2020-09-25 hellowOOOrld

A sequence of numbers is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.

For example, [1,7,4,9,2,5] is a wiggle sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, [1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.

Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

Examples:

Input: [1,7,4,9,2,5]
Output: 6
The entire sequence is a wiggle sequence.

Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].

Input: [1,2,3,4,5,6,7,8,9]
Output: 2

 

Follow up:
Can you do it in O(n) time?

思路：

用两个dp数组来维护状态的更新。

up[i]表示到i为止最后一个字符是上升的子序列长度。down[i]表示到i为止最后一个字符是下降的子序列长度。

由于摆动序列要求上升与下降交替进行。

故状态转移方程为

up[i] = max(up[i], down[j] + 1);

down[i] = max(down[i], up[j] + 1);

int wiggleMaxLength(vector<int>& nums)
     {
         int n = nums.size(),maxup=0,maxdown =0;
         vector<int>up(n + 1,0);
         vector<int>down(n + 1, 0);
         for (int i = 0; i <n;i++)
         {
             up[i] = 1;
             down[i] = 1;
             for (int j = 0; j < i;j++)
             {
                 if (nums[i]>nums[j])
                 {
                     up[i] = max(up[i], down[j] + 1);
                 }
                 if (nums[i]<nums[j])
                 {
                     down[i] = max(down[i], up[j] + 1);
                 }
             }
            // maxup = max(maxup, up[i]);
            // maxdown = max(maxdown, down[i]);
         }
         return max(up.back(), down.back());
     }

另外此题时间复杂度为O(n)的做法，暂时还没搞明白。。。