zoukankan      html  css  js  c++  java
  • [LintCode] Longest Increasing Subsequence 最长递增子序列

    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;
        }  
    };

    还有两种方法,(未完待续。。)

    参考资料:

    http://www.cnblogs.com/lishiblog/p/4190936.html

    http://blog.xiaohuahua.org/2015/01/26/lintcode-longest-increasing-subsequence/

  • 相关阅读:
    k8s使用私有镜像仓库
    spark client 配置lzo
    jvm系列(四):jvm调优-命令篇
    mysqldump 备份还原数据库
    df 卡死及ls无法查看文件
    记录一次服务器断电,直接进入救援模式
    nginx开机自启脚本
    mongodb启动关闭脚本
    mongo数据备份恢复
    centos 快速配置网络
  • 原文地址:https://www.cnblogs.com/grandyang/p/4891500.html
Copyright © 2011-2022 走看看