[LeetCode] 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]]

對(duì)于[i,j]按照i升序,j降序排序,然后按照j依次插入隊(duì)列。
矮個(gè)子對(duì)高個(gè)子沒有影響。

代碼:

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        if (people.length == 0) {
            return people;
        }
        List<int[]> l = new ArrayList<>();
        Arrays.sort(people, (o1, o2) -> o1[0] == o2[0] ? o1[1] - o2[1] : o2[0] - o1[0]);
        for (int i = 0; i < people.length; i++) {
            l.add(people[i][1], people[i]);
        }
        int[][] result = new int[people.length][people[0].length];
        for (int i = 0; i < l.size(); i++) {
            result[i] = l.get(i);
        }
        return result;
    }
}
?著作權(quán)歸作者所有,轉(zhuǎn)載或內(nèi)容合作請(qǐng)聯(lián)系作者
【社區(qū)內(nèi)容提示】社區(qū)部分內(nèi)容疑似由AI輔助生成,瀏覽時(shí)請(qǐng)結(jié)合常識(shí)與多方信息審慎甄別。
平臺(tái)聲明:文章內(nèi)容(如有圖片或視頻亦包括在內(nèi))由作者上傳并發(fā)布,文章內(nèi)容僅代表作者本人觀點(diǎn),簡(jiǎn)書系信息發(fā)布平臺(tái),僅提供信息存儲(chǔ)服務(wù)。

相關(guān)閱讀更多精彩內(nèi)容

友情鏈接更多精彩內(nèi)容