<p>Given an integer array <code>nums</code>, return <em>the length of the longest <strong>strictly increasing </strong></em><span data-keyword="subsequence-array"><em><strong>subsequence</strong></em></span>.</p>

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

<pre>
<strong>Input:</strong> nums = [10,9,2,5,3,7,101,18]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,0,3,2,3]
<strong>Output:</strong> 4
</pre>

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

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2500</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><b>Follow up:</b>&nbsp;Can you come up with an algorithm that runs in&nbsp;<code>O(n log(n))</code> time complexity?</p>

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> nums_sort;
        int i, j;
        int** ans;
        for (i=0;i<nums.size();i++) {
            nums_sort.push_back(nums[i]);
        }
        sort(nums_sort.begin(), nums_sort.end());
        i=1;
        while (i!=nums_sort.size()) {
            if (nums_sort[i-1] == nums_sort[i]) {
                nums_sort.erase(nums_sort.begin()+i);
            }
            else {
                i++;
            }
        }

        ans = new int *[nums_sort.size()+1];
        for (i=0;i<nums_sort.size()+1;i++) {
            ans[i] = new int [nums.size()+1];
            for (j=0;j<nums.size()+1;j++) {
                if (i==0 || j==0) {
                    ans[i][j]=0;
                }
                else {
                    if(nums[j-1] == nums_sort[i-1]) {
                        ans[i][j] = ans[i-1][j-1] + 1;
                    }
                    else {
                        if (ans[i-1][j] > ans[i][j-1]) {
                            ans[i][j] = ans[i-1][j];
                        }
                        else {
                            ans[i][j] = ans[i][j-1];
                        }
                    }
                }
            }
        }
        return ans[nums_sort.size()][nums.size()];
    }
};
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