zoukankan      html  css  js  c++  java
  • 【Lintcode】076.Longest Increasing Subsequence

    题目:

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

    You code should return the length of the LIS.

    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_increasing_subsequence

    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 [2, 4, 5, 7], return 4

    题解:

      For dp[i], dp[i] is max(dp[j]+1, dp[i]), for all j < i and nums[j] < nums[i].

    Solution 1 ()

    class Solution {
    public:
        int longestIncreasingSubsequence(vector<int> nums) {
            if (nums.empty()) {
                return 0;
            }
            vector<int> dp(nums.size(), 1);
            int res = 1;
            for (int i = 1; i < nums.size(); ++i) {
                for (int j = 0; j < i; ++j) {
                    if (nums[j] < nums[i]) {
                        dp[i] = max(dp[i], dp[j] + 1);
                    }
                }
                res = max(dp[i], res);
            }
            return res;
        }
    };

    Solution 2 ()

    class Solution {
    public:
        /**
         * @param nums: The integer array
         * @return: The length of LIS (longest increasing subsequence)
         */
        int longestIncreasingSubsequence(vector<int> nums) {
            vector<int> res;
            for(int i=0; i<nums.size(); i++) {
                auto it = std::lower_bound(res.begin(), res.end(), nums[i]);
                if(it==res.end()) res.push_back(nums[i]);
                else *it = nums[i];
            }
            return res.size();
        }
    };

    Solution 3 ()

    class Solution {
    public:
        int longestIncreasingSubsequence(vector<int> nums) {
            if (nums.empty()) {
                return 0;
            }
            vector<int> tmp;
            tmp.push_back(nums[0]);
            for (auto num : nums) {
                if (num < tmp[0]) {
                    tmp[0] = num;
                } else if (num > tmp.back()) {
                    tmp.push_back(num);
                } else {
                    int begin = 0, end = tmp.size();
                    while (begin < end) {
                        int mid = begin + (end - begin) / 2;
                        if (tmp[mid] < num) {
                            begin = mid + 1;
                        } else {
                            end = mid;
                        }
                    }
                    tmp[end] = num;
                }
            }
            return tmp.size();
        }
    };
  • 相关阅读:
    tomcat bug之部署应用的时候经常会发上startup failed due to previous errors
    maven编译项目理解
    MyReport报表引擎2.6.5.0新功能
    PHP入门-摘要表格处理问题
    EnumMap源代码阅读器
    MySQL几种方法的数据库备份
    工作日志2014-08-19
    Linux通过网卡驱动程序和版本号的信息
    JS于,子类调用父类的函数
    hdu 5007 水 弦
  • 原文地址:https://www.cnblogs.com/Atanisi/p/6883116.html
Copyright © 2011-2022 走看看