[leetcode]376. Wiggle Subsequence

题目链接：https://leetcode.com/problems/wiggle-subsequence/#/description

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

Follow up:

Can you do it in O(n) time?

思路: （贪心）
而为了能够在O(n)时间内解决可以考虑使用贪心法. 举个栗子: [1,17,5,10,13,15,10,5,16,8], 可以看到前两个[1, 17]确定了一个递增的序列, 而[17, 5]构成了一个递减序列, 所以到目前位置都正常. 到了[10, 13, 15]这里就有问题了, 他们和之前的5构成了一个递增序列, 而出于贪心的考虑, 必然是选择15是最优解, 因为这样给后面序列最大的选择空间.
对于接下来的[10, 5]都与之前的15构成递减区间, 同样道理我们选择5来构造这个序列. 所以一个基于贪心的算法大概就是这样了. 其时间复杂度为(n), 空间复杂度为O(1).

class Solution{
public:
int wiggleMaxLength(vector<int>& nums)
{
if (nums.size()<2)
return nums.size();
int len=nums.size(),ans=len,flag=0;
for(int i=1;i<len;i++)
{
if(nums[i]-nums[i-1]==0) ans--;
else if(nums[i]-nums[i-1]>0) flag==1?ans--:flag=1;
else if(nums[i]-nums[i-1]<0) flag==-1?ans--:flag=-1;
}
return ans;
}
};

class Solution{
public:
int wiggleMaxLength(vector<int>& nums)
{
if(nums.size()<2)
return nums.size();
int flag=0;
int res=1;
for(int i=1;i<nums.size();i++)
{
if(nums[i]>nums[i-1])
{
if(flag!=1)
{
res++;
flag=1;
}
}

else if(nums[i]<nums[i-1])
{
if(flag!=-1)
{
res++;
flag=-1;
}
}
}
return res;
}
};