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.
Examples:
Input: [1,7,4,9,2,5]
Output: 6
The entire sequence is a wiggle sequence.
Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
Input: [1,2,3,4,5,6,7,8,9]
Output: 2
摆动序列，就是一个序列，相邻两个数相减，组成的新序列是按正负正负正负...（第一个是正或者负，不是0）这么排的的序列。
求一个序列的最长摆动子序列的长度。可以把这个想象成坐标系里的一条折线，摆动序列的形状是一上一下的，有些情况可能是连续好几个点向上或者向下，只取最后一个点就变成摆动序列了。
思路是：找个标志位flag，记下前一个做差的符号，遍历数组，用本次的差和flag相比，本次的差是0就忽略掉，本次的差和之前一样也忽略掉（其实是相当于在子队列里保当前的最后元素，剔除前一个），和之前不一样就说明是子序列的元素。
把不是摆动序列通过删除元素改成摆动序列，上图，
1,2,3,三个节点删掉都能保证到它位置都是摆动序列，从贪心算法的思想考虑，应该删最后一个，所以遍历过程中是，是1和2比，删1，然后2和3比，删2。代码如下：
class Solution {
public:
int wiggleMaxLength(vector<int>& nums) {
int n = nums.size();
if(n <= 0)
return 0;
int res = 1;
int flag = 0;
for(int i = 1; i < n; i++)
{
int a = nums[i] - nums[i - 1];
if(a != 0)
a = a / abs(a);
else
continue;
if(a != flag)
res++;
flag = a;
}
return res;
}
};
