leetcode97 子数组最小值之和

/*class Solution
{
   public:
    using myType = pair<int, int>;
    struct cmp {
        bool operator()(const myType& m1, const myType& m2) const { return m1.second > m2.second; }
    };
    int sumSubarrayMins(vector<int>& A)
    {
        int len = A.size();
        if (len == 0) {
            return 0;
        }
        int sum = 0;
        // 双端队列插入的时候把最大值放队头
        for (int i = 1; i <= len; i++) {
            priority_queue<myType, vector<myType>, cmp> pq; // care1!只能这样写
            for (int k = 0; k < i; k++) {
                pq.push({k, A[k]});
            }
            cout << endl;
            sum += pq.top().second; // 初始窗口
            for (int k = i; k < len; k++) {
                pq.push({k, A[k]});
                while (pq.top().first < (k - i + 1)) {
                    pq.pop();
                }
                sum += pq.top().second;
            }
        }
        return sum;
    }
};*/
class Solution
{
   public:
    using myType = pair<int, int>;
    int sumSubarrayMins(vector<int>& A)
    {
        int len = A.size();
        if (len == 0) {
            return 0;
        }
        int sum = 0;
        // 双端队列插入的时候把最大值放队头
        auto f = [&A](int x, int y) { return A[x] > A[y]; };
        for (int i = 1; i <= len; i++) {
            priority_queue<int, vector<int>, decltype(f)> pq(f); // care2!只能这样写
            for (int k = 0; k < i; k++) {
                pq.push(k);
            }
            cout << endl;
            sum += A[pq.top()]; // 初始窗口
            for (int k = i; k < len; k++) {
                pq.push(k);
                while (pq.top() < (k - i + 1)) {
                    pq.pop();
                }
                sum += A[pq.top()];
            }
        }
        return sum;
    }
};
发布了33 篇原创文章 · 获赞 0 · 访问量 621

猜你喜欢

转载自blog.csdn.net/qq_28133013/article/details/103994002
今日推荐