# 406. Queue Reconstruction by Height (Medium)

https://leetcode.com/problems/queue-reconstruction-by-height/

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers `(h, k)`, where `h` is the height of the person and `k` is the number of people in front of this person who have a height greater than or equal to `h`. Write an algorithm to reconstruct the queue.

Note:
The number of people is less than 1,100.

Example

```Input:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

Output:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]
```

## Solutions

``````class Solution {

// This problem is difficult and tricky, not easy to get over it on my own. So the best option is to
// memorize the problem-solving idea and steps.

public int[][] reconstructQueue(int[][] people) {
if (people == null || people.length == 0 || people[0].length == 0)
return new int[0][0];

Arrays.sort(people, (a, b) -> {
if (b[0] == a[0]) {
// Same height, adopt ascending order by prior people
return a[1] - b[1];
}

// Descending order by people's height
return b[0] - a[0];
});

ArrayList<int[]> tmp = new ArrayList<>();

// Of same prior people, higher people must be situated after lower one, or else the lower
// one will has more prior in front of him than the higher, which brings about discrepancy against
// the given situation.

// In light of above rule, the pre-placed element on specific slot will be edge off by later people
// who comes with lower height and maintain the order without going against the given situation.
int len = people.length;
for (int i = 0; i < len; i++) {
}

int[][] ans = new int[people.length][2];
int i = 0;
for (int[] k : tmp) {
ans[i][0] = k[0];
ans[i++][1] = k[1];
}

return ans;
}
}
``````