Leetcode-406. Queue Reconstruction by Height

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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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]]

这个题目可以先把数组排序，然后插排就可以了。Your runtime beats 14.47% of java submissions.
public class Solution {
private int index = 0;
public int[][] reconstructQueue(int[][] people) {
Queue<int[]> queue = new PriorityQueue<int[]>(new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
if(o1[0] > o2[0]) return -1;
else if(o1[0] < o2[0]) return 1;
else{
if(o1[1] > o2[1]) return 1;
else if(o1[1] < o2[1]) return -1;
else return 0;
}
}
});
for(int [] item : people) queue.add(item);

List<int[]> result = new ArrayList<int[]>();
while(queue.peek() != null){
int bigNum = 0;
int[] item = queue.poll();

for(int i = 0; i < result.size(); i ++){
if(bigNum == item[1]) {result.add(i,item); break;}
if(result.get(i)[0] >= item[0]) bigNum ++;
}
if(bigNum == result.size()) result.add(item);

}
return result.toArray(new int[people.length][2]);

}

}