667. Beautiful Arrangement II.md

667. Beautiful Arrangement II

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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.

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Solution

• mine

class Solution {
    public int[] constructArray(int n, int k) {
        int[] result = new int[n];
        int start = 1;
        int end = n;
        for(int i = 0; i < n; i++){
            if(k % 2 == 0){
                result[i] = start;
                start++;
            }else{
                result[i] = end;
                end--;
            }
            if(k > 1){
                k--;
            }
        }
        return result;
    }
}

• the most vote

if you have n number, the maximum k can be n - 1;

if n is 9, max k is 8.

This can be done by picking numbers interleavingly from head and tail,

    // start from i = 1, j = n;
    // i++, j--, i++, j--, i++, j--
    i: 1   2   3   4   5
    j:   9   8   7   6
    out: 1 9 2 8 3 7 4 6 5
    dif:  8 7 6 5 4 3 2 1

Above is a case where k is exactly n - 1

When k is less than that, simply lay out the rest (i, j) in incremental

order(all diff is 1). Say if k is 5:

    i++ j-- i++ j--  i++ i++ i++ ...

    out: 1   9   2   8    3   4   5   6   7
    dif:   8   7   6   5    1   1   1   1 

• C++

class Solution {
public:
    vector<int> constructArray(int n, int k) {
        vector<int> res;
        for (int i = 1, j = n; i <= j; ) {
            if (k > 1) {
                res.push_back(k-- % 2 ? i++ : j--);
            }
            else {
                res.push_back(i++);
            }
        }

        return res;
    }
};

• C++ Compact

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

• Java

class Solution {
    public int[] constructArray(int n, int k) {
        int[] res = new int[n];
        for (int i = 0, l = 1, r = n; l <= r; i++)
            res[i] = k > 1 ? (k-- % 2 != 0 ? l++ : r--) : l++;
        return res;
    }
}