题目描述:
LeetCode 406. Queue Reconstruction by Height
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]]
题目大意:
假设有一队人随机站成一个圈。每个人通过一对整数(h, k)描述,其中h是其身高,k是站在他前面并且身高不低于他的人数。编写算法重构队列。
注意:
人数小于1100
解题思路:
首先选出k值为0且身高最低的人,记为hi, ki,将其加入结果集。
然后更新队列,若队列中人员的身高≤hi,则令其k值 - 1(需要记录原始的k值)。
循环直到队列为空。
Java代码:
public class Solution {
public int[][] reconstructQueue(int[][] people) {
int size = people.length;
LinkedList<int[]> list = new LinkedList<int[]>();
for (int i = 0; i < size; i++) {
list.add(new int[]{people[i][0], people[i][1], 0});
}
int ans[][] = new int[size][];
for (int i = 0; i < size; i++) {
Collections.sort(list, new Comparator<int[]>() {
public int compare (int[] a, int[] b) {
if (a[1] == b[1])
return a[0] - b[0];
return a[1] - b[1];
}
});
int[] head = list.removeFirst();
ans[i] = new int[]{head[0], head[1] + head[2]};
for (int[] p : list) {
if (p[0] <= head[0]) {
p[1] -= 1;
p[2] += 1;
}
}
}
return ans;
}
}
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