通过重新排列数组来最大化数组的绝对差之和

问题描述

以summation of the absolute differences between every two adjacent numbers is maximum

array =(1，2，7)安排是(1,7,2)

array=( 1, 2, 7)arrangement is (1,7,2)

sum=|1-7|+|7-2|=11

推荐答案

您可以在O(nlogn)中进行操作.请仔细考虑...

You can do so in O(nlogn).Think before...

1. 排序数组1 2 3 4 5

2. 选择left = 5 right = 5，这是最大值.我是1，而j是4，因此我给出了最大绝对差，所以

1. Sort the array 1 2 3 4 5

2. Select left=5 right=5 that is the maximum. i is 1 and j is 4 and i gives max absolute difference so

根据最大绝对差在其右侧或左侧添加.在此处以i表示右侧

Append on its right or left based upon max absolute difference.Here its on right with i

现在left = 1和right = 2和4与左差最大

Now left=1 and right=2 and 4 gives max difference with left so

最后，左和右的差异为3

Finally its same difference for left and right with 3

代码:-( C ++)

CODE:-(C++)

sort(a.begin(),a.end());
int l=a[a.size()-1]; //left
int r=l;             // right
int i=0,j=a.size()-2;
long long int sum=0;
while(i<j){
        int li=abs(l-a[i]),ri=abs(r-a[i]);
        int lj=abs(l-a[j]),rj=abs(r-a[j]);
        if(li>ri||lj>rj){ //left side
                if(li>lj){
                        sum+=li;
                        l=a[i++];
                }else{
                        sum+=lj;
                        l=a[j--];
                }
        }else{
                if(ri>rj){
                        sum+=ri;
                        r=a[i++];
                }else{
                        sum+=rj;
                        r=a[j--];
                }
        }
        //cout<<l<<"---"<<r<<"------"<<i<<"---"<<j<<"----------"<<sum<<endl;
}
sum+=MAX(abs(l-a[i]),abs(r-a[i]));
cout<<sum<<endl;

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