求最長擺動子序列。

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


這個題很有意思，有很多種解法，用來練練dp...然而還是看題解才會的 = =
 

一個普通dp代碼：

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <minmax.h>

using namespace std;

class Solution {
public:
	int wiggleMaxLength(vector<int>& nums) {
		if (nums.empty())
			return {};
		if (nums.size() < 2)
			return nums.size();

		vector<int> up(nums.size());
		vector<int> down(nums.size());

		for (int i = 1; i < nums.size(); i++)
			for (int j = 0; j < i; j++)
			{
				if (nums[i] > nums[j])
					up[i] = max(up[i], down[j] + 1);
				else if (nums[i] < nums[j])
					down[i] = max(down[i], up[j] + 1);
			}
		return 1 + max(up[nums.size() - 1], down[nums.size() - 1]);
	}
};

int main()
{
	vector<int> nums{ 1,17,5,10,13,15,10,5,16,8 };
	
	cout << Solution().wiggleMaxLength(nums) << endl;

	getchar();
	return 0;
}

　　運行結果：

7

　　時間復雜度O(n^2)，空間復雜度O(n)。

　　重點是這個優化的dp算法，時間復雜度O(n)，空間復雜度O(1)：

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <minmax.h>

using namespace std;

class Solution {
public:
	int wiggleMaxLength(vector<int>& nums) {
		if (nums.empty())
			return {};
		if (nums.size() < 2)
			return nums.size();

		int up = 1, down = 1;
		for (int i = 1; i < nums.size(); i++)
		{
			if (nums[i] > nums[i - 1])
				up = down + 1;
			else if (nums[i] < nums[i - 1])
				down = up + 1;
		}
		return max(up, down);
	}
};

int main()
{
	vector<int> nums{ 1,17,5,10,13,15,10,5,16,8 };
	
	cout << Solution().wiggleMaxLength(nums) << endl;

	getchar();
	return 0;
}

　　運行結果：

7

　　這個算法實在又簡潔又易懂。

LeetCode.376 - Wiggle Subsequence