Week13 - 376. Wiggle Subsequence

Week13 - 376. Wiggle Subsequence

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?

Credits:

Special thanks to @agave and @StefanPochmann for adding this problem and creating all test cases.

my solution:

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int size = nums.size();
        if (size == 0) {
            return 0;
        }
        vector<int> up(size, 0);
        vector<int> down(size, 0);
        up[0] = 1;
        down[0] = 1;
        for(int i=1; i<size; ++i){

            if (nums[i] > nums[i-1]) {
                up[i] = down[i-1] + 1;
                down[i] = down[i-1];
            }
            else if (nums[i] < nums[i-1]) {
                down[i] = up[i-1] + 1;
                up[i] = up[i-1];
            }
            else {
                up[i] = up[i-1];
                down[i] = down[i-1];
            }
        }
        return max(up[size-1], down[size-1]);
    }
};