[LintCode] Longest Increasing Subsequence 最长递增子序列-白红宇

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?

Example

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

Challenge

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

Clarification

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

我们先来看一种类似Brute Force的方法，这种方法会找出所有的递增的子序列，并把它们都保存起来，最后再找出里面最长的那个，时间复杂度为O(n²)，参见代码如下：

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

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