題目描述
Given an array a of length n, find another array, b, of length n such that:
for each i (1≤i≤n) MEX({b1, b2, …, bi})=ai.
The MEX of a set of integers is the smallest non-negative integer that doesn’t belong to this set.
If such array doesn’t exist, determine this.
Input
The first line contains an integer n (1≤n≤105) — the length of the array a.
The second line contains n integers a1, a2, …, an (0≤ai≤i) — the elements of the array a. It’s guaranteed that ai≤ai+1 for 1≤i<n.
Output
If there’s no such array, print a single line containing −1.
Otherwise, print a single line containing n integers b1, b2, …, bn (0≤bi≤106)
If there are multiple answers, print any.
Examples
inputCopy
3
1 2 3
outputCopy
0 1 2
inputCopy
4
0 0 0 2
outputCopy
1 3 4 0
inputCopy
3
1 1 3
outputCopy
0 2 1
Note
In the second test case, other answers like [1,1,1,0], for example, are valid.
題目分析
我們可以將a[]中沒有的數放入一個小根堆中，每次輸出堆頂元素，然後移除堆頂元素。如果a[]中某個元素已經沒有了，那麼就將該元素再放入堆中。
代碼如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=1e5+5;
int a[N];
bool st[N];
int main()
{
int n;
cin>>n;
priority_queue<int,vector<int>,greater<int> > heap;
for(int i=1;i<=n;i++)
{
cin>>a[i];
st[a[i]]=true; //記錄a[]中元素
}
for(int i=0;i<2*n;i++)
{
if(!st[i]) heap.push(i); //如果a[]中沒有某個元素，就將其放入堆中
}
for(int i=1;i<=n;i++)
{
cout<<heap.top()<<' '; //每次輸出堆頂元素
heap.pop();
if(a[i+1]!=a[i]) heap.push(a[i]); //當a[]中沒有相同的元素時，便可將a[i]放入堆中（a[]中的元素是單調遞增的）
}
return 0;
}