Question

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?

Solution

Greedy method. Use a stack to track the current wiggle sequence.

class Solution {
    public int wiggleMaxLength(int[] nums) {
        if(nums.length <= 1) return nums.length;

        int maxLen = 1;
        Stack<Integer> stack = new Stack<>();
        stack.push(nums[0]);
        boolean isIncrease = false; // initialize
        for(int i = 1; i < nums.length; i++) {
            if(nums[i] != stack.peek()) {
                if(stack.size() == 1) {
                    isIncrease = nums[i] > stack.peek();
                    stack.push(nums[i]);
                    maxLen = Math.max(maxLen, stack.size());
                }
                else {
                    if(isIncrease == (nums[i] < stack.peek())) {
                        isIncrease = !isIncrease;
                        stack.push(nums[i]);
                        maxLen = Math.max(maxLen, stack.size());
                    } else {
                        stack.pop();
                        stack.push(nums[i]);
                    }
                }
            }
        }
        return maxLen;
    }
}