README_EN.md

376. Wiggle Subsequence

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Description

A wiggle sequence is a sequence where 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 two or fewer 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) alternate between positive and negative.
• In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.

A subsequence is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Given an integer array nums, return the length of the longest wiggle subsequence of nums.

 

Example 1:

Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).

Example 2:

Input: nums = [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length.
One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).

Example 3:

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

 

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 1000

 

Follow up: Could you solve this in O(n) time?

Solutions

Dynamic programming.

Python3

class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        up = down = 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1]:
                up = max(up, down + 1)
            elif nums[i] < nums[i - 1]:
                down = max(down, up + 1)
        return max(up, down)

Java

class Solution {
    public int wiggleMaxLength(int[] nums) {
        int up = 1, down = 1;
        for (int i = 1; i < nums.length; ++i) {
            if (nums[i] > nums[i - 1]) {
                up = Math.max(up, down + 1);
            } else if (nums[i] < nums[i - 1]) {
                down = Math.max(down, up + 1);
            }
        }
        return Math.max(up, down);
    }
}

TypeScript

function wiggleMaxLength(nums: number[]): number {
    let up = 1,
        down = 1;
    for (let i = 1; i < nums.length; ++i) {
        let prev = nums[i - 1],
            cur = nums[i];
        if (cur > prev) {
            up = Math.max(up, down + 1);
        } else if (cur < prev) {
            down = Math.max(down, up + 1);
        }
    }
    return Math.max(up, down);
}

C++

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

Go

func wiggleMaxLength(nums []int) int {
	up, down := 1, 1
	for i := 1; i < len(nums); i++ {
		if nums[i] > nums[i-1] {
			up = max(up, down+1)
		} else if nums[i] < nums[i-1] {
			down = max(down, up+1)
		}
	}
	return max(up, down)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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