算法D52 | 动态规划13 | 300.最长递增子序列 674. 最长连续递增序列 718. 最长重复子数组

今天开始正式子序列系列，本题是比较简单的，感受感受一下子序列题目的思路。 

视频讲解：动态规划之子序列问题，元素不连续！| LeetCode：300.最长递增子序列_哔哩哔哩_bilibili

代码随想录

Python:

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        n = len(nums)
        if n==1: return 1
        dp = [1]*n
        result = 0
        for i in range(1, n):
            for j in range(i):
                if nums[i]>nums[j]:
                    dp[i] = max(dp[i], dp[j]+1)
            result = max(result, dp[i])
        return result

C++:

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

674. 最长连续递增序列 

本题相对于昨天的动态规划：300.最长递增子序列 最大的区别在于“连续”。 先尝试自己做做，感受一下区别  

视频讲解：动态规划之子序列问题，重点在于连续！| LeetCode：674.最长连续递增序列_哔哩哔哩_bilibili

代码随想录

Python:

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [1]*n
        result = 1
        for i in range(1, n):
            if nums[i]>nums[i-1]:
                dp[i] = dp[i-1]+1
            result = max(result, dp[i])
        return result

C++:

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

718. 最长重复子数组 

稍有难度，要使用二维dp数组了

视频讲解：动态规划之子序列问题，想清楚DP数组的定义 | LeetCode：718.最长重复子数组_哔哩哔哩_bilibili

代码随想录

Python:

class Solution:
    def findLength(self, nums1: List[int], nums2: List[int]) -> int:
        n, m = len(nums1), len(nums2)
        dp = [[0]*(m+1) for _ in range(n+1)]
        result = 0
        for i in range(1, n+1):
            for j in range(1, m+1):
                if nums1[i-1]==nums2[j-1]:
                    dp[i][j] = dp[i-1][j-1] + 1
                result = max(result, dp[i][j])
        return result
        

C++:

class Solution {
public:
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        vector<vector<int>> dp(nums1.size()+1, vector<int>(nums2.size()+1, 0));
        int result = 0;
        for (int i=1; i<=nums1.size(); i++) {
            for (int j=1; j<=nums2.size(); j++) {
                if (nums1[i-1]==nums2[j-1]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                result = max(result, dp[i][j]);
            } 
        }
        return result;
    }
};