376. Wiggle Subsequence

package LeetCode_376

/**
 * 376. Wiggle Subsequence
 * https://leetcode.com/problems/wiggle-subsequence/
 * 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. 1 <= nums.length <= 1000
2. 0 <= nums[i] <= 1000
 * */
class Solution {
    /**
     * solution: keep checking current different and prev different (up or down),
     * because we can find out rule by [6, -3, 5, -7, 3] and [16, -7, 3, -3, 6, -8],
     * prev diff:0
     * current diff:6
     * then
     * prev diff:6
     * current diff:-3
     * then
     * ...
     * Time:O(n), Space:O(1)
     * */
    fun wiggleMaxLength(nums: IntArray): Int {
        val size = nums.size
        if (size == 1) {
            return 1
        }
        var count = 1
        var prevDiff = 0
        var diff = 0
        for (i in 1 until size) {
            diff = nums[i] - nums[i - 1]
            //keep tracking down or up
            if (diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0) {
                count++
                prevDiff = diff
            }
        }
        return count
    }
}

 

posted @ 2021-08-01 23:20  johnny_zhao  阅读(33)  评论(0编辑  收藏  举报