376. Wiggle Subsequence

2021/8/1 23:35:55

本文主要是介绍376. Wiggle Subsequence,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

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
    }
}

 



这篇关于376. Wiggle Subsequence的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!


扫一扫关注最新编程教程