# 406 Queue Reconstruction by Height

### Problem:

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:

``````public class Solution {
public int[][] reconstructQueue(int[][] people) {
Arrays.sort(people, (new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
if (o1 == o2) {
return o1 - o2;
} else {
return o2 - o1;
}
}
}));
for(int[] cur : people){
}
return resultList.toArray(new int[people.length][]);
}
}
``````
``````public class Solution {
public int[][] reconstructQueue(int[][] people) {
Arrays.sort(people, (new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
if (o1 == o2) {
return o1 - o2;
} else {
return o2 - o1;
}
}
}));
for (int i = 1; i < people.length; ++i) {
int cnt = 0;
for (int j = 0; j < i; ++j) {
if (cnt == people[i]) {
int[] t = people[i];
for (int k = i - 1; k >= j; --k) {
people[k + 1] = people[k];
}
people[j] = t;
break;
}
if (people[j] >= people[i])
++cnt;
}
}
return people;
}
}
``````
``````class Solution {
public int[][] reconstructQueue(int[][] people) {
if (people.length == 0 || people.length == 0) {
return people;
}
int[][] res = new int[people.length][people.length];
boolean[] filled = new boolean[people.length];
Arrays.sort(people, new Comparator<int[]>(){
public int compare(int[] p1, int[] p2) {
if (p1 == p2) {
return p2 - p1;
}
else {
return p1 - p2;
}
}
});
for (int i = 0; i < people.length; i ++) {
int[] p = people[i];
int count = 0;
for (int j = 0; j < people.length; j ++) {
if (filled[j] == false) {
count ++;
if (count == p + 1) {
filled[j] = true;
res[j] = p;
res[j] = p;
break;
}
}
}
}
return res;
}
}
``````