There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. Given N distinct positive integers after a run of partition, could you tell how many elements could be the selected pivot for this partition?
For example, given N=5 and the numbers 1, 3, 2, 4, and 5. We have:
- 1 could be the pivot since there is no element to its left and all the elements to its right are larger than it;
- 3 must not be the pivot since although all the elements to its left are smaller, the number 2 to its right is less than it as well;
- 2 must not be the pivot since although all the elements to its right are larger, the number 3 to its left is larger than it as well;
- and for the similar reason, 4 and 5 could also be the pivot.
Hence in total there are 3 pivot candidates.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤105). Then the next line contains N distinct positive integers no larger than 109. The numbers in a line are separated by spaces.
Output Specification:
For each test case, output in the first line the number of pivot candidates. Then in the next line print these candidates in increasing order. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.
Sample Input:
5
1 3 2 4 5
Sample Output:
3
1 4 5
#include<bits/stdc++.h>
using namespace std;
const int N = 100000+5;
int n,c;
int a[N], l[N], r[N], ans[N];
int main(){
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &a[i]);
for(int i = 1; i < n; i++) l[i] = max(a[i-1], l[i-1]);
r[n-1] = (1<<31)-1;
for(int i = n-2; i >= 0; i--) r[i] = min(a[i+1], r[i+1]);
for(int i = 0; i < n; i++)
if(l[i] < a[i] && a[i] < r[i]) ans[c++] = a[i];
printf("%d\n", c);
for(int i = 0; i < c; i++){
if(i != 0) printf(" ");
printf("%d", ans[i]);
}
printf("\n");
return 0;
}
1、刚开始想到的思路是对数组排序，然后位置不变的元素是pivot，后来发现这只是必要条件
2、 #递推
l[i] 表示a[i] 左边最大的元素，r[i] 表示a[i] 右边最小的元素，a[i] 是pivot当且仅当 l[i] < a[i] && a[i] < r[i]。
3、当pivot个数是0时，第二行没有输出pivot，但必须输出换行。否则测试点2会报错。