Leetcode周赛189

暴力完事

class Solution {
public:
    int busyStudent(vector<int>& startTime, vector<int>& endTime, int queryTime) {
        // int a[1001];
        int ans=0;
        for(int i=0; i<startTime.size(); i++)
        {
            if(startTime[i]<=queryTime&&endTime[i]>=queryTime)
                    ans++;
        }
        return ans;
    }
};

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直接拆分排序组合

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class Solution {
public:
    static bool cmp(pair<int,string> &a,pair<int,string> &b){
        return a.second.length()==b.second.length()?a.first<b.first:a.second.length()<b.second.length();
    }
    string arrangeWords(string text) {
        vector<pair<int,string>> a;
        string ans="";
        int id=0;
        int len=text.length();
        int i=0;
        string temp="";
        while(i<len){
            temp="";
            while(i<len&&text[i]!=' '){
                temp+=text[i++];
            }
            i++;
            if(id==0)
                temp[0]=temp[0]+32;
            a.push_back(make_pair(id,temp));
            id++;
        }
        sort(a.begin(),a.end(), cmp);
        for(i=0; i<a.size()-1; i++)
            ans+=a[i].second+" ";
        ans += a[a.size()-1].second;
        ans[0] -=32;
        return ans;
    }
};

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先查总的独特的公司名，然后将每个人的收藏转换成bool数组，暴力就行

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class Solution {
public:
    bool in(bool cnt[100][1001], int i, int j, int len){
        for(int k=0; k<len; k++)
            if(cnt[i][k]==1&&cnt[j][k]==0)
                return false;
        return true;
    }
    vector<int> peopleIndexes(vector<vector<string>>& a) {
        bool cnt[100][1001];
        memset(cnt,0,sizeof cnt);
        set<string> se;
        for(auto i:a){
            for(auto ii:i)
                se.insert(ii);
        }
        int len=se.size();
        map<string, int> mp;
        int id=0;
        for(auto &i:se){
            mp[i]=id++;
        }
        for(int i=0; i<a.size(); i++){
            for(auto ii:a[i])
                cnt[i][mp[ii]]=1;
        }
        vector<int> ans;
        for(int i=0; i<a.size(); i++){
            bool flag=true;
            for(int j=0; j<a.size(); j++)
            {
                if(i==j||!in(cnt,i,j,len))
                    continue;
                else
                {
                    flag=false;
                    break;
                }
            }
            if(flag)
                ans.push_back(i);
        }
        return ans;
    }
};

没做出来，看了题解发现，最贪心的做法是让边界上起码有两个点
然后利用两个点和半径确定圆心，求当前的最大的覆盖，O(n^3)复杂度

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class Solution:
    def numPoints(self, points: List[List[int]], r: int) -> int:
        def dist2(p1: List[int], p2: List[int]) -> int:
            return (p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2
        
        def getCircle(p1: List[int], p2: List[int]) -> List[float]:
            mid = [(p1[0] + p2[0]) / 2, (p1[1] + p2[1]) / 2]
            angle = math.atan((p1[0] - p2[0]) / (p2[1] - p1[1])) if p2[1] - p1[1] else math.pi / 2
            d = math.sqrt(r2 - dist2(p1, mid))
            return [mid[0] + d * math.cos(angle), mid[1] + d * math.sin(angle)]
        
        n = len(points)
        ans = 1
        eps = 1e-6
        r2 = r * r
        for i in range(n):
            for j in range(i + 1):
                if dist2(points[i], points[j]) > 4 * r2:
                    continue
                center = getCircle(points[i], points[j])
                count = 0
                for k in range(n):
                    if dist2(points[k], center) < r2 + eps:
                        count += 1
                ans = max(ans, count)
        return ans