【wiggle-subsequence】leetcode-376

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

解题思路：其实只需要比较a[n+1]-a[n]和a[n]-a[n-1]的正负性即可，定义变量flag记录上一次的正负性，current记录上一个的值，count记录满足要求的总和，于是有以下代码：

class Solution(object):
    def wiggleMaxLength(self, nums):
        """ 
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        current = nums[0]
        count = 1 
        flag = None
        for num in nums[1:]:
            if flag == None:
                if num == current:
                    continue
                if num < current:
                    flag = False
                else:
                    flag = True
                count += 1
                current = num
            elif flag == True:
                if num < current:
                    flag = False
                    count += 1
                current = num
            else:
                if num > current:
                    flag = True
                    count += 1
                current = num
        return count

有几点需要注意的是：

1，初始情况是先向上走还是先向下走是未定的，不论哪种情况，都要将计数器+1

2，初始情况有可能是连续相等的值，此时情况仍未定