[LeetCode] Beautiful Arrangement II 优美排列之二
Given two integers n
and k
, you need to construct a list which contains n
different positive integers ranging from 1
to n
and obeys the following requirement:
Suppose this list is [a1, a2, a3, ... , an], then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] has exactly k
distinct integers.
If there are multiple answers, print any of them.
Example 1:
Input: n = 3, k = 1 Output: [1, 2, 3] Explanation: The [1, 2, 3] has three different positive integers ranging from 1 to 3, and the [1, 1] has exactly 1 distinct integer: 1.
Example 2:
Input: n = 3, k = 2 Output: [1, 3, 2] Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
Note:
- The
n
andk
are in the range 1 <= k < n <= 104.
这道题虽然也叫优美排列,但是貌似跟之前那道Beautiful Arrangement的关系不太大。这道题给我们了一个数字n和一个数字k,让找出一种排列方式,使得1到n组成的数组中相邻两个数的差的绝对值正好有k种。给了k和n的关系为k<n。那么我们首先来考虑,是否这种条件关系下,是否已定存在这种优美排列呢,我们用一个例子来分析,比如说当n=8,我们有数组:
1, 2, 3, 4, 5, 6, 7, 8
当我们这样有序排列的话,相邻两数的差的绝对值为1。我们想差的绝对值最大能为多少,应该是把1和8放到一起,为7。那么为了尽可能的产生不同的差的绝对值,我们在8后面需要放一个小数字,比如2,这样会产生差的绝对值6,同理,后面再跟一个大数,比如7,产生差的绝对值5,以此类推,我们得到下列数组:
1, 8, 2, 7, 3, 6, 4, 5
其差的绝对值为:7,6,5,4,3,2,1
共有7种,所以我们知道k最大为n-1,所以这样的排列一定会存在。我们的策略是,先按照这种最小最大数相邻的方法排列,没排一个,k自减1,当k减到1的时候,后面的排列方法只要按照生序的方法排列,就不会产生不同的差的绝对值,这种算法的时间复杂度是O(n),属于比较高效的那种。我们使用两个指针,初始时分别指向1和n,然后分别从i和j取数加入结果res,每取一个数字k自减1,直到k减到1的时候,开始按升序取后面的数字,参见代码如下:
解法一:
class Solution { public: vector<int> constructArray(int n, int k) { vector<int> res; int i = 1, j = n; while (i <= j) { if (k > 1) res.push_back(k-- % 2 ? i++ : j--); else res.push_back(i++); } return res; } };
下面这种方法是把上面的if...else的语句用三元操作符合并成了一句,看起来更加简洁了一些。
解法二:
class Solution { public: vector<int> constructArray(int n, int k) { vector<int> res; int i = 1, j = n; while (i <= j) { res.push_back(k > 1 ? (k-- % 2 ? i++ : j--) : i++); } return res; } };
类似题目:
参考资料:
https://discuss.leetcode.com/topic/101113/c-java-clean-code-4-liner