"Fun algorithm Problem" in the array, for all elements, find the next element larger than the current element

In the array, for all elements, find the next element larger than the current element

The meaning is, eg. Array is 3 1 2 5 4 6 7

So the result we need to get is 5 2 5 6 6 7-1

Here's how to fix it:

1. Violent match: O (n ^ 2) efficiency matches all elements in the past and is very inefficient
2. I've been prompted to think of an O (NLG n) Efficiency algorithm

Only need to scan the array once, we use a priority_queue to get the current smallest element

The data structures stored by Prority_queue are:

struct SC {    int key, flag;    BOOL operator < (const SC & a) const    {return A.key < key;}};

Key denotes the value of the element, and flag indicates the subscript, which is useful when assigning a value

Algorithm ideas:

Traversal once, if the current queue is empty, then push

Otherwise, if the minimum element of the first team is less than a[i]

Then assigning a value in the answer array gives the next element larger than the current element

If the queue is empty, break

My Source code:

#include <iostream>#include<cstring>#include<stack>#include<string>#include<ctime>#include<queue>#include<algorithm>using namespacestd;Const intMAXN = +;intA[MAXN], B[MAXN];structSC {intkey, Flag; BOOL operator< (ConstSC & A)Const    { returnA.key <key;}};intMain () {intI, J, K, U, V, n; memset (b,-1,sizeof(b)); CIN>>N;  for(i =0; I < n; ++i) Cin >>A[i]; Priority_queue<sc>que;  for(i =0; I < n; ++i) {if(Que.empty ()) {SC temp; Temp.key= A[i], Temp.flag =i;            Que.push (temp); Continue; }         for (;;) {            if(Que.empty ()) Break;//if pop to empty break;            if(Que.top (). Key <A[i]) {b[que.top (). Flag]=A[i];            Que.pop (); } Else  Break;        } SC temp; Temp.key=A[i]; Temp.flag=i;    Que.push (temp); }     for(i =0; I < n; ++i) {cout<< A[i] <<' '; } cout<<Endl;  for(i =0; I < n; ++i) {cout<< B[i] <<' '; } cout<<Endl; return 0;}

"Fun algorithm Problem" in the array, for all elements, find the next element larger than the current element