LeetCode 406.根据身高重建队列 | 解题思路及代码

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根据身高重建队列

原题：406. Queue Reconstruction by Height

Problem Description

There are \(n\) people, we want them line up in the following way.
Given a two-dimensional array: \(people[n][2]\), where \(people[i]=[h_i][k_i]\), \(h_i\) means the height of \(i\)-th person, \(k_i\) means there are \(k_i\) people who are the same height as person \(i\) or taller than person \(i\) stand in front of person \(i\). So there are actually \(k_i\) or more people stand in front of person \(i\).
According to the two-dimentional array, line them up. It's promised that them can be lined up.

example

people =\([[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]]\)
answer =\([[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]]\)

Ideas

Firstly, let's consider a simpler case: people's height differs. No two people are the same height.

No same height

Assume there are a array \(line[n]\). We want to put everyone in this line. When we put \(people[i]=[h_i][k_i]\), assuming no one taller than him has been lined. We need to reserve \(k_i\) spaces in front of him for those \(k_i\) people who are higher than him. So we put person \(i\) at the \(k_i+1\)-th empty position.
How to ensure "no one taller than him has been lined"? That's sample, just put them in the line in ascending order of height.

Now we consider how to handle people who have same height.

Same height

For person \(i\), \(j\), we assume their height are the same. Without loss of generality, we assume \(k_i<k_j\)(if they are equal, they can't be lined as required).
Then we have: person \(i\) must stand in front of person \(j\). Otherwise it won't meet the requirement. The question now is should we put person \(i\) first or person \(j\) first.
If we put person \(i\) first. When puting person \(j\), we should put him at the \(k_j+1\)-th position as explained above. Those \(k_j+1\) empty space are reserve for people has
the same height or higher, but we have put person \(i\) who has the same height as \(j\) and \(i\) is in front of \(j\), alreadly taken a place himself! We should only reserve \(k_j\) empty space. So puting \(i\) first is not optimal.
If we put person \(j\) first. When puting \(i\), \(i\) is in front of \(j\) and that just take the place \(j\) left for \(i\). So that's established.
So we put people who has same height in \(k\) descending order.

Algorithm

Sort \(people\) array according to \(h_i\) as the first keyword in ascending order and \(k_i\) as the second keyword in descending order.
Put the sorted \(people[i]\) in the \(k_i+1\)-th position in turn.

The detailed code is as follows

public class p406 {
    public int[][] reconstructQueue(int[][] people) {
        int num = people.length;
        Arrays.sort(people, new Comparator<int[]>() {
            public int compare(int[] t1, int[] t2) {
                return t1[0] == t2[0] ? t2[1] - t1[1] : t1[0] - t2[0];
            }
        });
        int[][] ans = new int[num][];
        for (int[] person : people) {
            int space = person[1] + 1;
            for (int i = 0; i < num; i++) {
                if (ans[i] == null) {
                    space--;
                }
                if (space == 0) {
                    ans[i] = person;
                    break;
                }
            }
        }
        return ans;
    }
}

Time&Space Complexity

Time complexity: \(O(\log n)\) for sort, \(O(n^2)\) for going through \(people\) array and find corresponding location for each person. So the total time complexity is \(O(n^2)\).
Space complexity: Sorting is implemented by recursive calls. So the time complexity is \(O(\log n)\).

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