Given a sequence of integers, find the longest increasing subsequence (LIS).

You code should return the length of the LIS.

Have you met this question in a real interview?

For [5, 4, 1, 2, 3], the LIS  is [1, 2, 3], return 3

For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4

Time complexity O(n^2) or O(nlogn)

What's the definition of longest increasing subsequence?

    * The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.  

    * https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

我們先來看一種類似Brute Force的方法，這種方法會找出所有的遞增的子序列，並把它們都儲存起來，最後再找出裡面最長的那個，時間複雜度為O(n²)，參見程式碼如下：

class Solution {
public:
    /**
     * @param nums: The integer array
     * @return: The length of LIS (longest increasing subsequence)
     
 
*/
    int longestIncreasingSubsequence(vector<int> nums) {
        vector<vector<int> > solutions;
        longestIncreasingSubsequence(nums, solutions, 0);
        int res = 0;
        for (auto &a : solutions) {
            res = max(res, (int)a.size());
        }
        return res;
    }
    void longestIncreasingSubsequence(vector<int> &nums, vector<vector<int> > &solutions, int curIdx) {
        if (curIdx >= nums.size() || curIdx < 0) return;
        int cur = nums[curIdx];
        vector<int> best_solution;
        for (int i = 0; i < curIdx; ++i) {
            if (nums[i] <= cur) {
                best_solution = seqWithMaxLength(best_solution, solutions[i]);
            }
        }
        vector<int> new_solution = best_solution;
        new_solution.push_back(cur);
        solutions.push_back(new_solution);
        longestIncreasingSubsequence(nums, solutions, curIdx + 1);
    }
    vector<int> seqWithMaxLength(vector<int> &seq1, vector<int> &seq2) {
        if (seq1.empty()) return seq2;
        if (seq2.empty()) return seq1;
        return seq1.size() < seq2.size() ? seq2 : seq1;
    }  
};

還有兩種方法，(未完待續。。)

參考資料：