Given an array A of integers, a ramp is a tuple (i, j) for which i < j and A[i] <= A[j].  The width of such a ramp is j - i.

Find the maximum width of a ramp in A.  If one doesn't exist, return 0.

Example 1:

Input: [6,0,8,2,1,5]
Output: 4
Explanation: 
The maximum width ramp is achieved at (i, j) = (1, 5): A[1] = 0 and A[5] = 5.
 

Example 2:

Input: [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation: 
The maximum width ramp is achieved at (i, j) = (2, 9): A[2] = 1 and A[9] = 1.

Note:

1. 2 <= A.length <= 50000
2. 0 <= A[i] <= 50000

題意在解釋裡說的很清楚了，就看怎麼想。

首先保證後面的數字比前面的是大於等於關係，這個排序可以搞定，然後到這個數為止，前面數字的位置裡最小的數是誰，減去就好了

struct P{
    int num;
    int pos;
}H[50002];
bool cmp(P a,P b){
    if(a.num == b.num){
        return a.pos<b.pos;
    }else{
        return a.num<b.num;
    }
}
class Solution {
public:

    int maxWidthRamp(vector<int>& A) {
        int len = A.size();
        
 
for(int i=0;i<len;i++){
            H[i].num = A[i];
            H[i].pos = i;
        }
        sort(H,H+len,cmp);
        int Min = H[0].pos;
        int sum = -1;
        int cum = 0;
        for(int i=1;i<len;i++){
            //cout<<H[i].pos<<" "<<Min<<endl;
            if(Min > H[i].pos){
                Min = H[i].pos;
            }else{
                
                cum = H[i].pos - Min;
                sum = max(sum,cum);
            }
            
        }
        if(sum == -1){
            sum = 0;
        }
        return sum;
    }
};