<p>A <strong>wiggle sequence</strong> is a sequence where 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 one element and a sequence with two non-equal elements are trivially wiggle sequences.</p>

<ul>
	<li>For example, <code>[1, 7, 4, 9, 2, 5]</code> is a <strong>wiggle sequence</strong> because the differences <code>(6, -3, 5, -7, 3)</code> alternate between positive and negative.</li>
	<li>In contrast, <code>[1, 4, 7, 2, 5]</code> and <code>[1, 7, 4, 5, 5]</code> are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.</li>
</ul>

<p>A <strong>subsequence</strong> is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.</p>

<p>Given an integer array <code>nums</code>, return <em>the length of the longest <strong>wiggle subsequence</strong> of </em><code>nums</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,7,4,9,2,5]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,17,5,10,13,15,10,5,16,8]
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are several subsequences that achieve this length.
One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7,8,9]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve this in <code>O(n)</code> time?</p>

class Solution {
public:
    // Analysis
    // method1，地推法，就是看和前面的值对比,然后再覆盖。
    // 其实也是一个DP算法拉。。就是假设前面[已知】
    int wiggleMaxLength(vector<int>& nums) {
      vector<pair<string, int>> v1; // head
      vector<pair<string, int>> v2; // tail

      for (int i = 0; i < nums.size(); i++) {
        v1.push_back(make_pair("head", 1));
        v2.push_back(make_pair("tail", 1));

        for (int j = 0; j < i; j++) {
            if (v1[j].first == "head") {
                if (nums[i] > nums[j]) {
                    int cnt = v1[j].second + 1;
                    if (cnt > v1[i].second) {
                        v1[i] = make_pair("tail", cnt);
                    }
                }
            } else {
                if (nums[i] < nums[j]) {
                    int cnt = v1[j].second + 1;
                    if (cnt > v1[i].second) {
                        v1[i] = make_pair("head", cnt);
                    } 
                }
            }

            if (v2[j].first == "head") {
                if (nums[i] > nums[j]) {
                    int cnt = v2[j].second + 1;
                    if (cnt > v2[i].second) {
                        v2[i] = make_pair("tail", cnt);
                    }
                }
            } else {
                if (nums[i] < nums[j]) {
                    int cnt = v2[j].second + 1;
                    if (cnt > v2[i].second) {
                        v2[i] = make_pair("head", cnt);
                    } 
                }
            }
        }
      }  

      int max_cnt = 1;
      for (int i = 0; i < v1.size(); i++) {
        cout  << v1[i].first << " " << v1[i].second << " ";
        if (max_cnt < v1[i].second) {
            max_cnt = v1[i].second;
        }
      }
      cout << endl;

    for (int i = 0; i < v2.size(); i++) {
        cout  << v2[i].first << " " << v2[i].second << " ";
        if (max_cnt < v2[i].second) {
            max_cnt = v2[i].second;
        }
      }
      cout << endl;
      return max_cnt;
    }
};
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