667. Beautiful Arrangement II(Java)

最新推荐文章于 2020-04-03 21:50:13 发布

Crystal_ting

最新推荐文章于 2020-04-03 21:50:13 发布

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分类专栏： leetcode Array

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本文介绍了一种算法，用于构造一个包含n个不同正整数的列表，这些整数范围从1到n，并且相邻元素的绝对差构成的新列表中恰好包含k个不同的整数。通过交错选取头部和尾部的数值，可以有效地解决这一问题。

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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 <= 10的四次方.

思路：

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--

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

class Solution {
    public int[] constructArray(int n, int k) {
        int[] res = new int[n];
        int max = n, min = 1;
        int i;
        for (i = 0; i < k; i ++) {
            if (i % 2 != 0) res[i] = max --;
            else res[i] = min ++;
        }
        if (i % 2 == 0) {
            for (int j = k; j < n; j ++) 
                res[j] = max --;
        } else {
            for (int j = k; j < n; j ++) 
                res[j] = min ++;
        }
        return res;
    }
}