Algorithm question check-in day52-dynamic programming | 300. Longest increasing subsequence, 674. Longest continuous increasing sequence, 718. Longest repeating subarray

300. The longest increasing subsequence - LeetCode
status: AC

It is possible to delete some elements in a subsequence without changing the original order. A double traversal is required for judgment. The code is as follows:

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

674. The longest continuous increasing sequence - LeetCode
status: AC

The increasing sequence is a continuous subsequence. A little trick is to put the final answer in a loop for updating. The code is as follows:

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

718. The longest repeated subarray - LeetCode
status: AC

An improved version of the previous question, the code is as follows:

class Solution {
public:
    int findLength(vector<int>& nums1, vector<int>& nums2) {
        int len1 = nums1.size(), len2 = nums2.size(), res = 0;
        vector<vector<int>> dp(len1+1, vector<int>(len2+1, 0));
        for(int i = 1; i <= len1; ++i){
            for(int j = 1; j <= len2; ++j){
                if(nums1[i-1] == nums2[j-1]){
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                res = max(res, dp[i][j]);   
            }
        }
        return res;
    }
};