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]]
一刷
题解：思路很巧妙
- 排序：高的在前，若一样高，k小的在前。
- 用LinkedList, 把<h, k>插入到位置k
因为，比如之前的位置是<h1, k1>, 现在出现了<h2, k2>， 那么必然有h2<h1, 那么不论插入到<h1, k1>之前还是之后，都不影响k1
class Solution {
public int[][] reconstructQueue(int[][] people) {
//pick up the tallest guy first
//when insert the next tall guy, just need to insert him into kth position
//repeat until all people are inserted into list
Arrays.sort(people, new Comparator<int[]>(){
public int compare(int[] a, int[]b){
if(a[0]!=b[0]) return b[0]-a[0];
else return a[1]-b[1];
}
});
List<int[]> res = new LinkedList<>();
for(int[] cur :people){
res.add(cur[1], cur);
}
return res.toArray(new int[people.length][]);//revert to int[][]
}
}
