Dynamic planning for solving problems (6)

300. The longest increasing subsequence

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

674. The longest continuous increasing sequence

summary ： Before the discontinuous increasing subsequence 0-i The states are related to , Continuously increasing subsequences are only related to the previous state

Be careful ：

i < nums.size() - 1;

class Solution {
    
public:
    int findLengthOfLCIS(vector<int>& nums) {
    
        if (nums.size() == 0) return 0;
        int result = 1;
        vector<int> dp(nums.size() ,1);
        for (int i = 0; i < nums.size() - 1; i++) {
    
            if (nums[i + 1] > nums[i]) {
     //  Continuous record 
                dp[i + 1] = dp[i] + 1;
            }
            if (dp[i + 1] > result) result = dp[i + 1];
        }
        return result;
    }
};

718. Longest repeating subarray

summary ：

dp[i][j] = dp[i - 1][j - 1] + 1;

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

Scrolling array Two dimensional to one dimensional array

   class Solution {
    
    public:
        int findLength(vector<int>& A, vector<int>& B) {
    
            vector<int> dp(vector<int>(B.size() + 1, 0));
            int result = 0;
            for (int i = 1; i <= A.size(); i++) {
    
                for (int j = B.size(); j > 0; j--) {
    
                    if (A[i - 1] == B[j - 1]) {
    
                        dp[j] = dp[j - 1] + 1;
                    } else dp[j] = 0; //  Note that when there is inequality here, there should be fu 0 The operation of 
                    if (dp[j] > result) result = dp[j];
                }
            }
            return result;
        }
    };

1143. Longest common subsequence

summary ：

dpi： The length is [0, i - 1] String text1 And length [0, j - 1] String text2 The longest common subsequence of is dpi

dpi = dpi - 1 + 1; dpi = max(dpi - 1, dpi);

error ：i <= text1.size(); j <= text2.size()

class Solution {
    
public:
    int longestCommonSubsequence(string text1, string text2) {
    
        vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, 0));
        for (int i = 1; i <= text1.size(); i++) {
    
            for (int j = 1; j <= text2.size(); j++) {
    
                if (text1[i - 1] == text2[j - 1]) {
    
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
    
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[text1.size()][text2.size()];
    }
};

1035. Disjoint lines

summary ：

ah It's exactly the same as the last one , I don't know another template .

class Solution {
    
public:
    int maxUncrossedLines(vector<int>& A, vector<int>& B) {
    
        vector<vector<int>> dp(A.size() + 1, vector<int>(B.size() + 1, 0));
        for (int i = 1; i <= A.size(); i++) {
    
            for (int j = 1; j <= B.size(); j++) {
    
                if (A[i - 1] == B[j - 1]) {
    
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
    
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[A.size()][B.size()];
    }
};

53. The largest subsequence and

summary ：

error ：dp[0] = nums[0];
int result = dp[0]; Initialize to 0

class Solution {
    
public:
    int maxSubArray(vector<int>& nums) {
    
        if (nums.size() == 0) return 0;
        vector<int> dp(nums.size());
        dp[0] = nums[0];
        int result = dp[0];
        for (int i = 1; i < nums.size(); i++) {
    
            dp[i] = max(dp[i - 1] + nums[i], nums[i]); //  State transition formula 
            if (dp[i] > result) result = dp[i]; // result  preservation dp[i] The maximum of 
        }
        return result;
    }
};

• Time complexity ：O(n)
• Spatial complexity ：O(n)

392. Judging subsequences

dpi Denotes the following i-1 String ending with s, And the following j-1 String ending with t, The length of the same subsequence is dpi.

class Solution {
    
public:
    bool isSubsequence(string s, string t) {
    
        vector<vector<int>> dp(s.size() + 1, vector<int>(t.size() + 1, 0));
        for (int i = 1; i <= s.size(); i++) {
    
            for (int j = 1; j <= t.size(); j++) {
    
                if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = dp[i][j - 1];
            }
        }
        if (dp[s.size()][t.size()] == s.size()) return true;
        return false;
    }
};

• Time complexity ：O(n × m)
• Spatial complexity ：O(n × m)