【LeetCode周赛7.31】题解

6134. 找到离给定两个节点最近的节点

class Solution {
public:
    int closestMeetingNode(vector<int>& e, int n1, int n2) {
        int n = e.size();
        vector< vector<int> > d (2, vector<int>(n,INT_MAX) );
        
        int dis = 0;
        vector<int> vis(n,0);
        for( int v = n1;v!=-1 && vis[v] ==0; v = e[v]){
            vis[v] = 1;
            d[0][v] = dis++; 
        }
        
        fill(vis.begin(),vis.end(),0);
        dis = 0;
        for( int v = n2;v!=-1 && vis[v] ==0; v = e[v]){
            vis[v] = 1;
            d[1][v] = dis++; 
        }

        int ret = INT_MAX;
        int v = -1;
        for(int i=0;i<n;++i){
            if(  max<int>(d[0][i],d[1][i]) < ret){
                ret = max<int>(d[0][i],d[1][i]);
                v = i;
            }
        }
        
        return v;
    }
};

使数组中所有元素都等于0

class Solution {
public:
    int minimumOperations(vector<int>& a) {
        sort(a.begin(),a.end());
        int n = unique(a.begin(),a.end())-a.begin();
        
        if(a[0]==0){
            --n;
        }
        return n;
    }
};

分组的最大数量

class Solution {
public:
    int maximumGroups(vector<int>& a) {
        
        int n = a.size();
        int i=1;
        for(i = 1; i*(i+1)/2 <= n ; ++i);
        return i-1;
    }
};

图中最长的环

class Solution {
public:
    int longestCycle(vector<int>& e) {
        int n = e.size();
        vector<int> v(n,0);
        int ret = -1;

        for(int i=0;i<n;++i){
            if( v[i]==0 ){
                
                unordered_map<int,int> rd;
                int index = 1;
                int x = i;
                while( x!=-1 ){

                    if( rd.count(x) > 0 ){
                        ret = max<int>(ret, index-rd[x] );
                    }
                    
                    if(v[x] == 1){
                        break;;
                    }

                    v[x] = 1;
                    rd[x] = index;
                    
                    x = e[x];
                    ++index;
                   
                }
            }
        }
        return ret;
    }
};