README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0376.Wiggle%20Subsequence/README_EN.md	Greedy Array Dynamic Programming

376. Wiggle Subsequence

中文文档

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 one element and a sequence with two non-equal elements are trivially wiggle sequences.

• 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

Solution 1: Dynamic Programming

We define $f[i]$ as the length of the wiggle sequence ending at the $i$th element with an upward trend, and $g[i]$ as the length of the wiggle sequence ending at the $i$th element with a downward trend. Initially, $f[0] = g[0] = 1$ because when there is only one element, the length of the wiggle sequence is $1$. Initialize the answer as $1$.

For $f[i]$, where $i \geq 1$, we enumerate $j$ in the range $[0, i)$, if $nums[j] < nums[i]$, it means that $i$ can be appended after $j$ to form an upward wiggle sequence, then $f[i] = \max(f[i], g[j] + 1)$; if $nums[j] > nums[i]$, it means that $i$ can be appended after $j$ to form a downward wiggle sequence, then $g[i] = \max(g[i], f[j] + 1)$. Then we update the answer to $\max(f[i], g[i])$.

Finally, we return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Python3

class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 1
        f = [1] * n
        g = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    f[i] = max(f[i], g[j] + 1)
                elif nums[j] > nums[i]:
                    g[i] = max(g[i], f[j] + 1)
            ans = max(ans, f[i], g[i])
        return ans

Java

class Solution {
    public int wiggleMaxLength(int[] nums) {
        int n = nums.length;
        int ans = 1;
        int[] f = new int[n];
        int[] g = new int[n];
        f[0] = 1;
        g[0] = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    f[i] = Math.max(f[i], g[j] + 1);
                } else if (nums[j] > nums[i]) {
                    g[i] = Math.max(g[i], f[j] + 1);
                }
            }
            ans = Math.max(ans, Math.max(f[i], g[i]));
        }
        return ans;
    }
}

C++

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int n = nums.size();
        int ans = 1;
        vector<int> f(n, 1);
        vector<int> g(n, 1);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    f[i] = max(f[i], g[j] + 1);
                } else if (nums[j] > nums[i]) {
                    g[i] = max(g[i], f[j] + 1);
                }
            }
            ans = max({ans, f[i], g[i]});
        }
        return ans;
    }
};

Go

func wiggleMaxLength(nums []int) int {
	n := len(nums)
	f := make([]int, n)
	g := make([]int, n)
	f[0], g[0] = 1, 1
	ans := 1
	for i := 1; i < n; i++ {
		for j := 0; j < i; j++ {
			if nums[j] < nums[i] {
				f[i] = max(f[i], g[j]+1)
			} else if nums[j] > nums[i] {
				g[i] = max(g[i], f[j]+1)
			}
		}
		ans = max(ans, max(f[i], g[i]))
	}
	return ans
}

TypeScript

function wiggleMaxLength(nums: number[]): number {
    const n = nums.length;
    const f: number[] = Array(n).fill(1);
    const g: number[] = Array(n).fill(1);
    let ans = 1;
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            if (nums[i] > nums[j]) {
                f[i] = Math.max(f[i], g[j] + 1);
            } else if (nums[i] < nums[j]) {
                g[i] = Math.max(g[i], f[j] + 1);
            }
        }
        ans = Math.max(ans, f[i], g[i]);
    }
    return ans;
}