376. Wiggle Subsequence ----M !!!!!!!!!~~~~~~~~~

 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?

A:

哎，看着很难，可是我真正开了番茄，逼迫自己想的时候。

其实还是可以25分钟内做出来的。 反而直接bug free

核心就是： 每次找到寻找下一个bigger, (or 下一个smaller)  如果找到，则皆大欢喜。

否则，就调整前一个pre(which should be less than its left, )

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class Solution { public: int wiggleMaxLength(vector<int>& nums) { if(nums.size()==0) return 0; return max(helper(nums,1,false), helper(nums,1,true)); } private: int helper(vector<int>& nums, int start, bool isNextBigger){ int n = nums.size(); int pre = nums[start-1]; int res = 1; for(int i= start; i< n; i++){ int cur = nums[i]; if(isNextBigger){ if(cur > pre){ isNextBigger = ! isNextBigger; pre = cur; res++; }else{ // cur <= pre if( i-2 < 0 || cur < nums[i-2] ){ pre = cur; } } }else{ // we want cur should be smaller than previous if(cur < pre){ isNextBigger = ! isNextBigger; pre = cur; res++; }else{ // we want a smaller value, but get a bigger one if(i-2<0 || cur > nums[i-2] ){ pre = cur; } } } } return res; } };

Mistakes：

唯一的，就是没有检查 空输入，因为我们直接从nums[1] 开始的。

其实还有一个情况，是， 没有检查nums[0] 是否为INT_MIN or INT_MAX.