907. Sum of Subarray Minimums

思路：A[i]是最低值的次数 = （左边连续的大于A[i]的个数+1）*（右边连续大于等于A[i]的个数+1）

class Solution {
public:
    int sumSubarrayMins(vector& A) {
        long long sum = 0;
        int n = A.size();
        vector l(n,-1),r(n,n),st;
        for(int i = 0;iA[i])st.pop_back();
            l[i] = st.empty()?-1:st.back();
            st.push_back(i);
        }
        st.clear();
        for(int i = n-1;i>-1;i--){
            while(!st.empty()&&A[st.back()]>=A[i])st.pop_back();//注意这里
            r[i] = st.empty()?n:st.back();
            st.push_back(i);
        }
        for(int i =0;i