contest 47的第一题也是很扯的一道题,思路很简单,但是题目出的很无聊,后两题的思路和解法都属于正常范畴, 这篇文章主要分析667的思路。
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 and k are in the range 1 <= k < n <= 104.
这题简单点说就是求1-n构成的数组的一个序,这个序要满足两两之间元素差的绝对值(相同的值算一种情况)构成的数组大小为k。返回满足条件的数组序列。 这题我拿到的第一反应是回溯法,尝试解了一下发现复杂度太高,然后就尝试着去找规律,发现是很简单的一个规律题。首先我们确定的一点:元素差最多的个数是n-1个,这个n-1的构成也很容易发现,较大的数和较小的数交替形成的序列就满足要求。例如,我们假设n= 6,k = 5,那么这个序列就是 6 1 5 2 4 3 形成的k个元素差为: 5 4 3 2 1 (反向也是可以的 即 1 6 2 5 3 4)若k不等于n-1,我们只需要按上述规律形成满足k-1的序列,剩余序列按递减序即可(剩余的差值都为1)。 假设n = 6,k = 4,我们得到的序列即是: 1 6 2 5 4 3。 按照上述思路可以写出很多AC解,其中一种如下:
class Solution { public: vector<int> constructArray(int n, int k) { vector<int> mVec; // 最大n-1 for(int i = 1,j = n;i<=j;) { if(k>1) { mVec.push_back(k--%2?i++:j--); }else { mVec.push_back(i++); } } return mVec; } };这种思路利用k来做插入的判断条件,其中i,j的位置可互换。