Leetcode 406. Queue Reconstruction by Height 排队重构 解题报告

1 解题思想

2 原题

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]]

3 AC解

public class Solution {
private void swap(int[][] people,int a,int b){
int t1=people[a][0],t2=people[a][1];
people[a][0] = people[b][0];
people[a][1] = people[b][1];
people[b][0] = t1;
people[b][1] = t2;

}
public int[][] reconstructQueue(int[][] people) {
//java 排序不方便，我这里就直接暴力排序了
//让身高按照降序排列，高的在前面，同身高的情况下让要求前面人数人少的在前面，就可以了
int n = people.length;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(people[i][0] < people[j][0])
swap(people,i,j);
else if(people[i][0] == people[j][0] && people[i][1] > people[j][1])
swap(people,i,j);
}
}
//按照顺序插入
List<Integer> la = new ArrayList<Integer>();
List<Integer> lb = new ArrayList<Integer>();
for(int i=0;i<n;i++){
la.add(people[i][1],people[i][0]);
lb.add(people[i][1],people[i][1]);
}
for(int i=0;i<n;i++){
people[i][0]=la.get(i);
people[i][1]=lb.get(i);
}
return people;

}
}