Daily LeetCode 376. Wiggle Subsequence

https://leetcode.com/problems/wiggle-subsequence/

Medium

问题描述：

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.

Example 1:

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

Example 2:

Input: [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].

Example 3:

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

Follow up:
Can you do it in O(n) time?

题目分析及代码：

题目给出了一个称作wiggle sequence的定义，这是一个摆动序列，满足nums[i-1]<nums[i]>nums[i+1]，我们需要从给定输入中找出一个最长的子摆动序列。

方法1:

对于数组中的每个元素，可能有三种状态：

• 后一个数大于这个数，此时记做up
• 后一个数小于这个数，此时记做down
• 后一个数等于这个数

我们维护两个数组，来记录当前情况下（index=i）wiggle sequence的最长长度：

• 后一个数大于当前数时，这个序列是“向上摆”的，想要确保该序列是摆动序列，我们需要确保当前数与前一个数的关系是“向下摆”，即当前数小于前一个数。此时更新up数组，up[i]=down[i-1]+1；同时，down数组在当前位置的值与先前相同，不做更新，down[i]=down[i-1]
• 后一个数小于当前数时，与上述情况刚好相反，此时，up[i]=up[i-1]，down[i]=up[i-1]+1
• 两数相等时，不做任何更新操作，up[i]=up[i-1], down[i]=down[i-1]

def wiggleMaxLength(nums):
    up = [1 for i in range(len(nums))]
    down = [1 for i in range(len(nums))]

    for i in range(1, len(nums)):
        if nums[i] > nums[i - 1]:
            up[i] = down[i - 1] + 1
            down[i] = down[i - 1]
        elif nums[i] < nums[i - 1]:
            down[i] = up[i - 1] + 1
            up[i] = up[i - 1]
        else:
            up[i], down[i] = up[i - 1], down[i - 1]

    return max(up[-1], down[-1])

方法2:

贪心策略，维护up和down两个变量，均初始化为1，遍历数组，当前元素大于前一个元素时，up = down + 1；当前元素小于前一个元素时，down = up + 1。

与动态规划思想相同，但节省了空间。

def wiggleMaxLength(nums):
    up, down = 1, 1
    for i in range(1, len(nums)):
        if nums[i] > nums[i - 1]:
            up = down + 1
        elif nums[i] < nums[i - 1]:
            down = up + 1

    return min(len(nums), max(up, down))