leetcode 376. Wiggle Subsequence

376. 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

Follow up:

Can you do it in O(n) time?
1、先用常规方式写
class Solution {
public:
int wiggleMaxLength(vector<int>& nums)
{
if (nums.empty()) return 0;
vector<int> flag (nums.size(), 0);//前一个到当前是递增就为1，递减就是-1
vector<int> vt (nums.size(), 1); //记录连续到当前这一个最大多少
for (int i = 1; i < nums.size(); i++)
{
for (int j = i - 1; j >= 0; j--)
{
if (nums[i] > nums[j] && (flag[j] == -1 || flag[j] == 0))
{
flag[i] = 1;
vt[i] = vt[j] + 1;
break;
}
else if (nums[i] < nums[j] && (flag[j] == 1 || flag[j] == 0))
{
flag[i] = -1;
vt[i] = vt[j] + 1;
break;
}
}
}
return *max_element(vt.begin(), vt.end());
}
};

2、改良，其实可以只与前一个比较。
class Solution {
public:
int wiggleMaxLength(vector<int>& nums)
{
if (nums.empty()) return 0;
int ret = 1;
int flag = 0; //1:需要一个大的 -1:需要一个小的 0:都可以
for (int i = 1; i < nums.size(); i++)
{
if (nums[i] > nums[i - 1] && (flag == 0 || flag == 1))
{
flag = -1; //之后需要一个小的
ret ++;
}
else if (nums[i] < nums[i - 1] && (flag == 0 || flag == -1))
{
flag = 1; //之后需要一个小的
ret ++;
}
}
return ret;
}
};