ACM-Sushi

2019.7.24

Arkady invited Anna for a dinner to a sushi restaurant. The restaurant is a bit unusual: it offers n pieces of sushi aligned in a row, and a customer has to choose a continuous subsegment of these sushi to buy.

The pieces of sushi are of two types: either with tuna or with eel. Let's denote the type of the i-th from the left sushi as ti, where ti=1 means it is with tuna, and ti=2 means it is with eel.

Arkady does not like tuna, Anna does not like eel. Arkady wants to choose such a continuous subsegment of sushi that it has equal number of sushi of each type and each half of the subsegment has only sushi of one type. For example, subsegment [2,2,2,1,1,1] is valid, but subsegment [1,2,1,2,1,2] is not, because both halves contain both types of sushi.

Find the length of the longest continuous subsegment of sushi Arkady can buy.

Input
The first line contains a single integer n (2≤n≤100000) — the number of pieces of sushi.

The second line contains n integers t1, t2, ..., tn (ti=1, denoting a sushi with tuna or ti=2, denoting a sushi with eel), representing the types of sushi from left to right.

It is guaranteed that there is at least one piece of sushi of each type. Note that it means that there is at least one valid continuous segment.

Output
Print a single integer — the maximum length of a valid continuous segment.

Examples
Input
7
2 2 2 1 1 2 2
Output
4
Input
6
1 2 1 2 1 2
Output
2
Input
9
2 2 1 1 1 2 2 2 2
Output
6

我的思路：

用一个一维数组a存储寿司的种类，再另开一个f数组,用一个for循环把a数组遍历一遍，如果第i个和第i+1个数不同，就把a数组该元素的下标加一标记进f数组里；再用一个for循环计算f数组里相邻两项元素的差值，将最大的差值存入一个变量max里面，最后输出两倍max即为所求。

#include <iostream>
using namespace std;
int main()
{
    int i,n,l1,l2;
    int a[100005]={0};
    int f[100005]; 
    f[0]=0;  //用来计算第一个输入的寿司种类的个数
    cin >> n;
    int max=-1, j=1;
    int m;
    for(i=0; i<n; i++)
    {
        cin >> a[i];
    }
    
    for(i=0; i<n-1; i++)
    {
        if(a[i] != a[i+1])
        {
            f[j]=i+1;
            j++;
        }
        f[j]=n;
    }
    
    if(j>1)
    {
        for(int i=1; i<j; i++)
        {
            l1=f[i]-f[i-1];
            l2=f[i+1]-f[i];
            m = (l1>=l2)?l2:l1;
            if(m>max)
                max=m;
        }
        cout << max*2;
    }
    else
        cout << "0";
    return 0;
}