LeetCode oj 406. Queue Reconstruction by Height(优先队列）

406. Queue Reconstruction by Height

 

QuestionEditorial Solution

 My Submissions

• Total Accepted: 1911
• Total Submissions: 3604
• Difficulty: Medium

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]]

给你一个二维数组，a[x][0]代表身高，a[x][1]代表前面有几个比a[x][0]高或者身高相等的，按照这个输出排序后的数组

这题卡了好久，本来思路是对的，就是优先队列加一个插入操作，结果看别人的错误题解看的自己卡了好几个小时，我还以为有不用插入操作的做法呢= =

说一下思路，先对二维数组排序，h不同的h大的在前面，h相同的k小的在前面，这个应该很好理解。排序的话可以用快排或者优先队列，我选择了优先队列，

复杂度是O(nlogn)，注意插入的时候要标记一下这个元素的位置，不然插入之后找不回去了。然后将队列中的元素挨个取出，按照k的大小依次往结果数组

里插入原数组元素，这里就需要到刚才那个标记的元素的位置了，我用的ArrayList实现的这一步骤。

public class Solution {    public class People{int h;int k;int index;public People(int h,int k,int index){this.h = h;this.k = k;this.index = index;}}public class compare implements Comparator<People>{public int compare(People x,People y){if(x.h == y.h){return x.k - y.k;}else{return y.h - x.h;}}}    public int[][] reconstructQueue(int[][] people) {        PriorityQueue<People> queue = new PriorityQueue<People>(new compare());List <Integer> link = new ArrayList<Integer>();int len = people.length;int ans[][] = new int [len][2];for(int i=0;i<len;i++){queue.add(new People(people[i][0],people[i][1],i));}while(!queue.isEmpty()){People p = queue.poll();link.add(p.k,p.index);}for(int i=0;i<len;i++){ans[i][0] = people[link.get(i)][0];ans[i][1] = people[link.get(i)][1];}return ans;    }}