请教大家一道题：wiggle subsequence

376. Wiggle Subsequence 这道题是dynamic programming的题。我看了官方的解法，一和二都懂了，但是对于第三种线性时间的解法存在疑惑。
这个解法核心是这样的：

Any element in the array could correspond to only one of the three possible states:

1. up position, it means nums[i] > nums[i-1]
2. down position, it means nums[i] < nums[i-1]
3. equals to position, nums[i] == nums[i-1]
The updates are done as:

If nums[i] > nums[i-1], that means it wiggles up. The element before it must be a down position. So up[i] = down[i-1] + 1, down[i] remains the same as down[i-1]. If nums[i] < nums[i-1], that means it wiggles down. The element before it must be a up position. So down[i] = up[i-1] + 1, up[i] remains the same as up[i-1]. If nums[i] == nums[i-1], that means it will not change anything becaue it didn't wiggle at all. So both down[i] and up[i] remain the same as down[i-1] and up[i−1].

我不明白的为什么1只比较了nums[i] > nums[i-1], 难道说nums[i]一定要连接上nums[i-1]咯？可是不一定呀，比方说有可能一个数组中有连续递增的几个数，nums[i]和nums[i-2]连起来也可以呀。我看了网上其他解释也没有讲这一点。

不知道有木有理解对楼主的疑问~ 这题是算subsequence的长度，所以nums[i]的状态是从nums[i-1]的状态转移过来的，不能直接从nums[i-2]跳到nums[i].