PAT-B1045-快速排序

#include <cstdio>
#include <algorithm>
using namespace std;

//思路：对于A[i],max(A[1..i-1] ,A[i]) = A[i]; min(A[i+1..N-1] ,A[i]) = A[i] 即为主元 
int A[100010], ans[100010], minright[100010];
int maxn = 1000000001;
int main(){
	int N;
	scanf("%d",&N);
	for(int i=0; i<N; i++){
		scanf("%d", &A[i]);
	}
	
	int maxleft = 0, minr = maxn, count = 0, index = 0;
	//用minright[j]数组记录A[j+1...N-1]的最小值
	for(int j = N-1; j>=0; j--){
		if(j == N-1) minright[j] = maxn;
		else{
			minr = min(minr, A[j+1]);
			minright[j] = minr;
		}	
	}
	//实现前面讲的思路
	for(int i=0; i<N; i++){
		if(i == 0){
			if(A[i] < minright[i]){
				count++;
				ans[index++] = A[i];
			}
		}
		else if(i == N-1){
			if(A[i] > maxleft){
				count++;
				ans[index++] = A[i];
			}
		}
		else{
			maxleft = max(maxleft,A[i-1]);
			if(A[i] > maxleft && A[i] < minright[i]){
				count++;
				ans[index++] = A[i];
			}
		}
	}
	printf("%d\n",count);
    if(count == 0){
		printf("\n");
		return 0;
	}//没有这个if,有个测试样例格式出错 
    sort(ans,ans+count);
	for(int i=0; i<count; i++){
		printf("%d",ans[i]);
		if(i != count-1) printf(" ");
	}
    return 0;
}