300. Longest Increasing Subseque

Problem:

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

Idea:

Dynamic Programming

solution1

// Author: Huahua
// Running time: 29 ms
class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if (nums.empty()) return 0;
        int n = nums.size();
        auto f = vector<int>(n, 1);
        for (int i = 1; i < n; ++i)
            for (int j = 0; j < i; ++j)
                if (nums[i] > nums[j])
                    f[i] = max(f[i], f[j] + 1);
        return *max_element(f.begin(), f.end());
    }
};

solution2

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;        
        f_ = vector<int>(n, 0);        
        int ans = 0;
        for (int i = 0; i < n; ++i)
            ans = max(ans, LIS(nums, i));
        return ans;
    }
private:
    vector<int> f_;
    // length of LIS ends with nums[r]
    int LIS(const vector<int>& nums, int r) {
        if (r == 0) return 1;
        if (f_[r] > 0) return f_[r];
        int ans = 1;
        for(int i = 0; i < r; ++i)
            if (nums[r] > nums[i])
                ans = max(ans, LIS(nums, i) + 1);
        f_[r] = ans;
        return f_[r];
    }
};

solution3

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

代码来源

https://zxi.mytechroad.com/blog/dynamic-programming/leetcode-300-longest-increasing-subsequence/