376. Wiggle Subsequence -Medium

最新推荐文章于 2020-09-01 18:16:25 发布

Euadvancer

最新推荐文章于 2020-09-01 18:16:25 发布

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分类专栏： LeetCode-动态规划文章标签： python leetcode 动态规划

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本文介绍了一种使用动态规划解决寻找最长摇摆子序列的问题的方法。摇摆序列是指序列中相邻元素差值正负交替出现的情况。文章提供了一个Python实现的例子，并详细解释了其背后的逻辑。

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Question

A sequence of numbers is called a wiggle sequence if 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 fewer than two 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) are alternately positive and negative. In contrast, [1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.

Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

如果一个连续序列的数字之间的差值按照正负交替出现，这个序列称为摇摆序列。摇摆序列的第一个差值既可以是正也可以是负。两个元素的序列也称为摇摆序列。给出一个正整数序列，返回最长摇摆子序列的长度。（子序列允许通过删除一些元素得到剩余的序列）

Example

Input: [1,7,4,9,2,5]
Output: 6
The entire sequence is a wiggle sequence.

Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].

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

Solution

这道题用dp解。思路如下：每当新加入一个序列中的元素，只会有3种状态

nums[i] > nums[i - 1]，即向上摇摆
nums[i] < nums[i - 1]，即向下摇摆
nums[i] = nums[i - 1]，不摇摆

那么我们需要每次加入新元素时分别记录向上摇摆up和向下摇摆down的最长序列的长度

如果向上摇摆，up = down + 1
如果向下摇摆，down = up + 1
如果不摇摆，跳过

class Solution(object):
    def wiggleMaxLength(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0: return 0
        up, down = 1, 1
        for index_n in range(1, len(nums)):
            if nums[index_n] > nums[index_n - 1]:
                up = down + 1
            elif nums[index_n] < nums[index_n - 1]:
                down = up + 1
        return max(down, up)