題目：
鏈接：https://leetcode.com/problems/wiggle-subsequence/
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
題目的意思很簡單，在數組中找到一個最長的子序列，使得這個序列的任意三個連續數字之間的差是相反的。也就是子序列中的數字的差是正負(或是負正交替出現的)。
源碼：
class Solution {
public:
bool is_same(int pre,int next){
if (pre > 0 && next > 0)return false;
else if (pre <0 && next <0)return false;
else return true;
}
int wiggleMaxLength(vector<int>& nums) {
vector<int>num;
int size = nums.size();
for (int i = 0; i < size; i++){
if(num.size()==0)num.push_back(nums[i]);
else{
if (nums[i] != *--num.end())num.push_back(nums[i]);
}
}
int size2 = num.size();
if (size2 == 0)return 0;
if (size2 == 1)return 1;
vector<int>diff(size2 - 1, 0);
vector<int>re(size2-1,0);
for (int i = 1; i < size2; i++)diff[i - 1] = num[i] - num[i - 1];
re[0] = 1;
for (int i = 1; i < size2 - 1; i++){
if (is_same(diff[i - 1],diff[i]))re[i] = re[i - 1] + 1;
else re[i] = re[i - 1];
}
return *--re.end() + 1;
}
};Submission Result: Accepted More Details
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在處理數組時，先進行相鄰元素的去重，去重之後，數組裏的任意兩個元素的差都不會是0，如果第diff[i]和diff[i-1]符號相同，diff[i]=diff[i-1]表示到第i個差的時候，前面正負交替
負正交替出現的差的最大個數和到第i-1個差的時候相同，否則加1,；