题目地址: https://leetcode.com/problems/wiggle-subsequence/description/
题目描述：
Asequence 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.
Forexample, [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:
Canyou do it in O(n) time?
题目大意
如果一个数组里面，相邻的两个数字的差是正负交替的，那么认为这个是波动序列。求输入的数组里面最长的波动序列长度。
解题方法
明显的DP问题，本来的想法是用个二维DP，可是提交了几遍只通过了部分测试用例。才去看的别人的两个DP数组的解法。
定义了一个记录递增的DP数组inc，一个记录递减的DP数组dec，这两个DP数组分别保存的是开头元素是递增、递减的最长波动序列长度。对于每个位置，从头遍历，如果当前的元素比前面的元素大，应该更新递增数组，否则，如果比前面的数字小，那么应该更新递减数组。
时间复杂度是O(N^2)，空间复杂度是O(N).
class Solution(object):
def wiggleMaxLength(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
if n <= 1:
return n
inc, dec = [1] * n, [1] * n
for x in range(n):
for y in range(x):
if nums[x] > nums[y]:
inc[x] = max(inc[x], dec[y] + 1)
elif nums[x] < nums[y]:
dec[x] = max(dec[x], inc[y] + 1)
return max(inc[-1], dec[-1])
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其实不需要从头遍历，只需要知道前面元素对应的最长递增和递减数组即可。
时间复杂度是O(N)，空间复杂度是O(N).
class Solution(object):
def wiggleMaxLength(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
if n <= 1:
return n
inc, dec = [1] * n, [1] * n
for x in range(1, n):
if nums[x] > nums[x - 1]:
inc[x] = dec[x - 1] + 1
dec[x] = dec[x - 1]
elif nums[x] < nums[x - 1]:
inc[x] = inc[x - 1]
dec[x] = inc[x - 1] + 1
else:
inc[x] = inc[x - 1]
dec[x] = dec[x - 1]
return max(inc[-1], dec[-1])
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简单分析代码就可以看出，每个元素都只和它之前的元素相关，因此，只需要使用两个变量即可。
时间复杂度是O(N)，空间复杂度是O(1).
class Solution(object):
def wiggleMaxLength(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
if n <= 1:
return n
inc, dec = 1, 1
for x in range(1, n):
if nums[x] > nums[x - 1]:
inc = dec + 1
elif nums[x] < nums[x - 1]:
dec = inc + 1
return max(inc, dec)
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参考资料：
https://leetcode.com/articles/wiggle-subsequence/ http://www.cnblogs.com/grandyang/p/5697621.html http://bookshadow.com/weblog/2016/07/21/leetcode-wiggle-subsequence/
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