【2022-05-22-第294场单周赛】

总结

Q1.字母在字符串中的百分比

RATING:1162

模拟

class Solution {
public:
    int percentageLetter(string s, char letter) {
        int n = s.size();
        int h[26] = {0};
        for(auto i : s) ++h[i - 'a'];
        int mx = 0;
        return 100 * h[letter - 'a'] / n;
    }
};

Q2.装满石头的背包的最大数量

RATING:1249

优先队列存每个背包剩余空间。

class Solution {
public:
    int maximumBags(vector<int>& c, vector<int>& r, int ar) {
        int ret = 0;
        priority_queue<int, vector<int>, greater<int>> q;
        for(int i = 0; i < c.size();++i){
            q.push(c[i] - r[i]);
        }
        while(ar > 0 && !q.empty()){
            if(ar >= q.top()){
                ar -= q.top();
                q.pop();
                ++ret;
            }
            else break;
        }
        return ret;
    }
};

也可排序+前缀和+二分。

class Solution {
public:
    int maximumBags(vector<int>& c, vector<int>& rk, int k) {
        int n = c.size();
        long long rest[n], ps[n + 1];
        ps[0] = 0;
        for(int i = 0; i < n; ++i) rest[i] = c[i] - rk[i];
        sort(rest, rest + n);
        for(int i = 0; i < n; ++i) ps[i + 1] = ps[i] + rest[i];
        int l = -1, r = n;
        while(l + 1 < r){
            int mid = l + r >> 1;
            if(ps[mid + 1] <= k) l = mid;
            else r = mid;
        }
        return l + 1;
    }
};

Q3.表示一个折线图的最少线段数

RATING:1681

要注意浮点数除法精度问题，一劳永逸的方法就是用乘法。

class Solution {
public:
    int minimumLines(vector<vector<int>>& s) {
        if(s.size() == 1) return 0;
        sort(s.begin(), s.end());
        /*vector<double> k;
        for(int i = 1; i < s.size(); ++i){
            k.push_back((double)((double)s[i][1] - s[i - 1][1]) / ((double)s[i][0] - s[i - 1][0]));
        }*/
        int ret = 1;
        for(int i = 2; i < s.size(); ++i){
            long long x1 = s[i - 2][0], y1 = s[i - 2][1], x2 = s[i - 1][0], y2 = s[i - 1][1], x3 = s[i][0], y3
 = s[i][1];
            if((y3 - y2) * (x1 - x2) != (y1 -y2) * (x3 - x2)) ++ret;
        }
        return ret;
    }
};

Q4.巫师的总力量和

RATING:2621

单调栈+前缀和之前缀和。（回头补）