PAT 甲级 刷题日记|A 1101 Quick Sort (25 分)

1101 Quick Sort 25分

单词积累

• partition 划分 分割

• pivot 枢纽 中心

• distinct positive integers 不同的正整数

题目

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

思路分析

暴力遍历的方法一定不可取，这里采取的是以时间换空间的思想：将每一位的左侧最大值和右侧最小值保存在数组中，可以通过两次遍历就实现（正序一次，逆序一次）。

活用递推 by 晴神笔记.png

还有一个小细节，存储时第i个位置的数组，存储的是i前的最大值，和i后的最小值。不要包含i，若包含i，后续判断时就需要访问i-1和i+1的位置，可能超出数组范围，处理较麻烦。

代码

#include 
using namespace std;

const int maxn = 100000;
int num[maxn];
int smaller[maxn];
int bigger[maxn];
int res[maxn];

int main () {
    int N;
    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf("%d", &num[i]);
        if (i > 0) {
            smaller[i] = max(num[i-1],smaller[i-1]);
        } else {
            smaller[i] = 0;
        } 
    }
    for (int i = N - 1; i >= 0; i--) {
        if (i < N-1) {
            bigger[i] = min(num[i+1], bigger[i+1]);
        } else {
            bigger[i] = INT_MAX;
        } 
    }
    int ans = 0;
    for (int i = 0; i < N; i++) {
        if (num[i] < bigger[i] && num[i] > smaller[i]) {
            res[ans] = num[i];
            ans++;
        } 
    }
    sort(res, res+ans);
    cout< 0) cout<