leetcode-----193场周赛

链接：https://leetcode-cn.com/contest/weekly-contest-193

一维数组的动态和

代码

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        vector<int> sum(nums.size());
        sum[0] = nums[0];
        for (int i = 1; i < nums.size(); i ++ )
            sum[i] = sum[i - 1] + nums[i];
        return sum;
    }
};

不同整数的最少数目

代码

class Solution {
public:
    int findLeastNumOfUniqueInts(vector<int>& arr, int k) {
        unordered_map<int, int> map;
        for (auto x: arr) map[x]++;

        vector<int> cnt;

        for (auto x: map) cnt.push_back(x.second);
        sort(cnt.begin(), cnt.end());

        int res = cnt.size();
        for (auto c: cnt) {
            if (k >= c) {
                k -= c;
                res--;
            } else {
                break;
            }
        }
        return res;
    }
};

制作 m 束花所需的最少天数

代码

class Solution {
public:

    int get(int l, int r, int k) {
        return (r - l + 1) / k;
    }

    int minDays(vector<int>& bs, int m, int k) {
        vector<pair<int, int>> q;
        int n = bs.size();
        for (int i = 0; i < bs.size(); i ++ ) q.push_back({bs[i], i + 1});
        
        vector<int> l(n + 2), r(n + 2);
        sort(q.begin(), q.end());
        int sum = 0;
        for (auto x : q) {
            int i = x.second;
            if (l[i - 1] && r[i + 1]) {
                sum = sum - get(l[i - 1], i - 1, k) - get(i + 1, r[i + 1], k) + get(l[i - 1], r[i + 1], k);
                r[l[i - 1]] = r[i + 1];
                l[r[i + 1]] = l[i - 1];
            } else if (l[i - 1]) {
                sum = sum - get(l[i - 1], i - 1, k) + get(l[i - 1], i, k);
                r[l[i - 1]] = i;
                l[i] = l[i - 1];
            } else if (r[i + 1]) {
                sum = sum - get(i + 1, r[i + 1], k) + get(i, r[i + 1], k);
                r[i] = r[i + 1];
                l[r[i + 1]] = i;
            } else {
                sum = sum + get(i, i, k);
                r[i] = l[i] = i;
            }

            if (sum >= m) return x.first;
        }

        return -1;
    }
};