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]]
Solution1：排序后插入
思路：
按照身高降序排列，同身高按前面人数量增序.
然后新建一个数组，遍历之前排好序的数组，根据每个元素的第二个数字[前面人数量]，将其插入到res数组中对应的位置
Time Complexity: O(N^2) Space Complexity: O(N)
Solution Code：
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[]>(){
@Override
public int compare(int[] o1, int[] o2){
return o1[0]!=o2[0]?-o1[0]+o2[0]:o1[1]-o2[1];
}
});
// java 8 lambda
// Arrays.sort(people, (a, b) -> a[0] != b[0] ? b[0] - a[0] : 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][]);
}
}
Solution2：
思路：
Time Complexity: O() Space Complexity: O()
