Saicode.com: flip array

给定一个长度为n的整数数组a，元素均不相同，问数组是否存在这样一个片段，只将该片段翻转就可以使整个数组升序排列。其中数组片段[l,r]表示序列a[l], a[l+1], ..., a[r]。原始数组为

a[1], a[2], ..., a[l-2], a[l-1], a[l], a[l+1], ..., a[r-1], a[r], a[r+1], a[r+2], ..., a[n-1], a[n]，

将片段[l,r]反序后的数组是

a[1], a[2], ..., a[l-2], a[l-1], a[r], a[r-1], ..., a[l+1], a[l], a[r+1], a[r+2], ..., a[n-1], a[n]。

 

Input   The first line of data is an integer: n (1≤n≤105), which indicates the length of the array. The second line of data is n integers a [1], a [2], ..., a [n] (1≤a [i] ≤109).	Sample input 4 2 1 3 4
Output   Output "yes" if it exists; otherwise output "no" without quotation marks.	Sample output   yes
Time limit C / C ++ language: 1000MS Other languages: 3000MS	Memory limit C / C ++ language: 65536KB Other languages: 589824KB

 Code directly:

#include <iostream>
using namespace std;

bool rotateArray(int *ar, int n){
    if(n == 1) return true;
    int pos1 = -1, pos2 = -1;
    bool flag = false;
    for (int i = 0; i < n - 1 ; ++i) {
        if(ar[i] > ar[i+ 1] && !flag){
            pos1 = i;
            flag = true;
        }
        if(ar[i] < ar[i +1] && flag){
            pos2 = i;
            break;
        }
    }
    for (int j = 0; j <= (pos2 - pos1)/2; ++j) {
        int tmp = ar[pos1 + j];
        ar[pos1 + j] = ar[pos2 - j];
        ar[pos2 - j] = tmp;
    }
    for (int k = 0; k < n - 1; ++k) {
        if(ar[k] > ar[k + 1]) return false;
    }
    return true;
}

int main() {
    int n;
    cin >> n;
    int ar[n];
    for (int i = 0; i < n; ++i) {
        cin >> ar[i];
    }
    if(rotateArray(ar, n)) cout << "yes" << endl;
    else cout << "no" << endl;
    return 0;
}