6136. 算术三元组的数目

class Solution {
    
    public:
        int arithmeticTriplets(vector<int>& nums, int diff) {
    
            int n=nums.size();
            int res=0;
            for(int i=0;i<n;i++)
            {
    
                for(int j=i+1;j<n;j++)
                {
    
                    for(int k=j+1;k<n;k++)
                    {
    
                        if(nums[j]-nums[i]==diff&&nums[k]-nums[j]==diff)res++;
                    }
                }
            }
            return res;
        }
    };

6139. 受限条件下可到达节点的数目

class Solution {
    
public:
    int reachableNodes(int n, vector<vector<int>>& edges, vector<int>& restricted) {
    
        int res=1;
        unordered_map<int,vector<int>>g;
        for(auto t:edges)
        {
    
            int a=t[0],b=t[1];
            g[a].push_back(b);
            g[b].push_back(a);
        } 
        unordered_set<int>s={
    restricted.begin(),restricted.end()};
        queue<int>q;
        vector<bool>st(n+10,false);
        q.push(0);
        st[0]=true;
        while(q.size())
        {
    
            auto t=q.front();
            q.pop();
            for(auto x:g[t])
            {
    
                if(!st[x]&&!s.count(x))
                {
    
                    res++;
                    q.push(x);
                    st[x]=true;
                }
            }
        }
        return res;
    }
};

6137. 检查数组是否存在有效划分

class Solution {
    
public:
    bool validPartition(vector<int>& nums) {
    
        int n=nums.size();
        vector<bool>f(n+10,false);
        f[0]=true;
        for(int i=1;i<n;i++)
        {
    
            if (f[i - 1] && nums[i] == nums[i - 1] ||
                i > 1 && f[i - 2] && (nums[i] == nums[i - 1] && nums[i] == nums[i - 2] ||
                                      nums[i] == nums[i - 1] + 1 && nums[i] == nums[i - 2] + 2))
            f[i+1]=true;
        }
        return f[n];
    }
};

6138. 最长理想子序列

class Solution {
    
public:

    int longestIdealString(string s, int k) {
    
        int n=s.size();
        vector<int>f(26,0);
        for(int i=0;i<s.size();i++)
        {
    
            int c=s[i]-'a';
            int m=0;
            for(int j=max(0,c-k);j<=min(25,c+k);j++)
                m=max(m,f[j]);
            f[c]=m+1;
        }
        int res=0;
        for(int i=0;i<26;i++)res=max(res,f[i]);
        return res;
    }
};