Formatted question description: https://leetcode.ca/all/667.html

# 667. Beautiful Arrangement II (Medium)

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:

1. The n and k are in the range 1 <= k < n <= 104.

Related Topics:
Array

Similar Questions:

## Solution 1.

// OJ: https://leetcode.com/problems/beautiful-arrangement-ii/

// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> constructArray(int n, int k) {
vector<int> ans(n);
for (int i = 0, val = 1, diff = k, dir = 1; i < n; ++i) {
ans[i] = val;
if (diff == 0) {
dir = 0;
val = 2 + k;
diff = 1; // any non-zero value
} else if (dir) {
val += diff * dir;
--diff;
dir = -dir;
} else ++val;
}
return ans;
}
};


Or

// OJ: https://leetcode.com/problems/beautiful-arrangement-ii/

// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> constructArray(int n, int k) {
vector<int> ans(n, 1);
int i = 1;
for (int val = 1, diff = k, dir = 1; diff; ++i) {
val += diff * dir;
--diff;
dir = -dir;
ans[i] = val;
}
for (int val = 2 + k; i < n; ++i, ++val) ans[i] = val;
return ans;
}
};


Java

class Solution {
public int[] constructArray(int n, int k) {
int[] sorted = new int[n];
for (int i = 0; i < n; i++)
sorted[i] = i + 1;
int[] arrangement = new int[n];
int low = 0, high = k;
for (int i = 0; i <= k; i++) {
if (i % 2 == 0)
arrangement[i] = sorted[low++];
else
arrangement[i] = sorted[high--];
}
for (int i = k + 1; i < n; i++)
arrangement[i] = sorted[i];
return arrangement;
}
}