667. Beautiful Arrangement II

找规律

1，2，... , n 乱序排列，相邻数据的绝对差最多有n-1种

比如1，2，3，4，5对应于 1 5 2 4 3

class Solution {
public:vector<int> constructArray(int n, int k) {vector<int> res;int l=1,r=k+1;while(l<=r){res.push_back(l++);if(l<=r){res.push_back(r--);}}for(int i=k+2;i<=n;++i){res.push_back(i);}return res;}
};

python代码

class Solution(object):def constructArray(self, n, k):""":type n: int:type k: int:rtype: List[int]"""res=[]l,r=1,k+1while l<=r:res.append(l)l+=1if l<=r:res.append(r)r-=1for i in xrange(k+2,n+1):res.append(i)return res

python 代码

class Solution(object):def constructArray(self, n, k):""":type n: int:type k: int:rtype: List[int]"""res=range(1,n-k)for i in range(0,k+1):if i%2==0:#print "{} is even".format(i)res.append(n-k+i//2)else:#print "{} is odd".format(i)res.append(n-i//2)return res;

答案

Approach #2: Construction [Accepted]

Intuition

When k = n-1, a valid construction is [1, n, 2, n-1, 3, n-2, ....]. One way to see this is, we need to have a difference of n-1, which means we need 1 and n adjacent; then, we need a difference of n-2, etc.

Also, when k = 1, a valid construction is [1, 2, 3, ..., n]. So we have a construction when n-k is tiny, and when it is large. This leads to the idea that we can stitch together these two constructions: we can put [1, 2, ..., n-k-1] first so that n is effectively k+1, and then finish the construction with the first "k = n-1"method.

For example, when n = 6 and k = 3, we will construct the array as [1, 2, 3, 6, 4, 5]. This consists of two parts: a construction of [1, 2] and a construction of [1, 4, 2, 3] where every element had 2 added to it (i.e. [3, 6, 4, 5]).

Algorithm

As before, write [1, 2, ..., n-k-1] first. The remaining k+1 elements to be written are [n-k, n-k+1, ..., n], and we'll write them in alternating head and tail order.

When we are writing the ithith element from the remaining k+1, every even ii is going to be chosen from the head, and will have value n-k + i//2. Every odd ii is going to be chosen from the tail, and will have value n - i//2.

class Solution(object):def constructArray(self, n, k):ans = list(range(1, n - k))for i in range(k+1):if i % 2 == 0:ans.append(n-k + i//2)else:ans.append(n - i//2)return ans

转载于:https://www.cnblogs.com/learning-c/p/9266391.html

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667. Beautiful Arrangement II

Approach #2: Construction [Accepted]

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