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

最新推荐文章于 2024-07-27 15:12:39 发布

转载最新推荐文章于 2024-07-27 15:12:39 发布 · 139 阅读

本文介绍了一种解决最长递增子序列(LIS)问题的方法，通过寻找所有可能的递增子序列并选取最长的一个来解决问题。示例中给出了具体的实现代码及两个示例，该方法的时间复杂度为O(n^2)。

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<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;
    }  
};