376, Wiggle Subsequence
I Problem
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 <= 10000 <= nums[i] <= 1000
Follow up
Could you solve this in O(n) time?
Related Topics
- Greedy
- Array
- Dynamic Programming
II Solution
Approach 1: Greedy
pub fn wiggle_max_length(nums: Vec<i32>) -> i32 {
let len = nums.len();
if len <= 1 {
return len as i32;
}
let mut prev_diff = nums[1] - nums[0];
let mut res = if prev_diff == 0 { 1 } else { 2 };
for i in 2..len {
let diff = nums[i] - nums[i - 1];
if (diff > 0 && prev_diff <= 0) || (diff < 0 && prev_diff >= 0) {
prev_diff = diff;
res += 1;
}
}
res
}public int wiggleMaxLength(int[] nums) {
int len = nums.length;
if (len <= 1) {
return len;
}
int prevDiff = nums[1] - nums[0];
int res = prevDiff == 0 ? 1 : 2;
for (int i = 2; i < len; i++) {
int diff = nums[i] - nums[i - 1];
if ((diff > 0 && prevDiff <= 0) || (diff < 0 && prevDiff >= 0)) {
prevDiff = diff;
res++;
}
}
return res;
}Approach 2: Dynamic Programming
pub fn wiggle_max_length(nums: Vec<i32>) -> i32 {
let len = nums.len();
if len <= 1 {
return len as i32;
}
let (mut up, mut down) = (vec![0; len], vec![0; len]);
(up[0], down[0]) = (1, 1);
for i in 1..len {
if nums[i - 1] < nums[i] {
up[i] = std::cmp::max(up[i - 1], down[i - 1] + 1);
down[i] = down[i - 1];
} else if nums[i - 1] > nums[i] {
up[i] = up[i - 1];
down[i] = std::cmp::max(up[i - 1] + 1, down[i - 1]);
} else {
up[i] = up[i - 1];
down[i] = down[i - 1];
}
}
std::cmp::max(up[len - 1], down[len - 1])
}public int wiggleMaxLength(int[] nums) {
int len = nums.length;
if (len <= 1) {
return len;
}
int[] up = new int[len], down = new int[len];
up[0] = down[0] = 1;
for (int i = 1; i < len; i++) {
if (nums[i - 1] < nums[i]) {
up[i] = Math.max(up[i - 1], down[i - 1] + 1);
down[i] = down[i - 1];
} else if (nums[i - 1] > nums[i]) {
up[i] = up[i - 1];
down[i] = Math.max(up[i - 1] + 1, down[i - 1]);
} else {
up[i] = up[i - 1];
down[i] = down[i - 1];
}
}
return Math.max(up[len - 1], down[len - 1]);
}Approach 3: Optimized Dynamic Programming
pub fn wiggle_max_length(nums: Vec<i32>) -> i32 {
let len = nums.len();
if len <= 1 {
return len as i32;
}
let (mut up, mut down) = (1, 1);
for i in 1..len {
if nums[i - 1] < nums[i] {
up = std::cmp::max(up, down + 1);
} else if nums[i - 1] > nums[i] {
down = std::cmp::max(up + 1, down);
}
}
std::cmp::max(up, down)
}public int wiggleMaxLength(int[] nums) {
int len = nums.length;
if (len <= 1) {
return len;
}
int up = 1, down = 1;
for (int i = 1; i < len; i++) {
if (nums[i - 1] < nums[i]) {
up = Math.max(up, down + 1);
} else if (nums[i - 1] > nums[i]) {
down = Math.max(up + 1, down);
}
}
return Math.max(up, down);
}