原文链接: https://leetcode-cn.com/problems/group-the-people-given-the-group-size-they-belong-to

英文原文

There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1.

You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3.

Return a list of groups such that each person i is in a group of size groupSizes[i].

Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.

 

Example 1:

Input: groupSizes = [3,3,3,3,3,1,3]
Output: [[5],[0,1,2],[3,4,6]]
Explanation: 
The first group is [5]. The size is 1, and groupSizes[5] = 1.
The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].

Example 2:

Input: groupSizes = [2,1,3,3,3,2]
Output: [[1],[0,5],[2,3,4]]

 

Constraints:

• groupSizes.length == n
• 1 <= n <= 500
• 1 <= groupSizes[i] <= n

中文题目

有 n 位用户参加活动，他们的 ID 从 0 到 n - 1，每位用户都 恰好 属于某一用户组。给你一个长度为 n 的数组 groupSizes，其中包含每位用户所处的用户组的大小，请你返回用户分组情况（存在的用户组以及每个组中用户的 ID）。

你可以任何顺序返回解决方案，ID 的顺序也不受限制。此外，题目给出的数据保证至少存在一种解决方案。

 

示例 1：

输入：groupSizes = [3,3,3,3,3,1,3]
输出：[[5],[0,1,2],[3,4,6]]
解释： 
其他可能的解决方案有 [[2,1,6],[5],[0,4,3]] 和 [[5],[0,6,2],[4,3,1]]。

示例 2：

输入：groupSizes = [2,1,3,3,3,2]
输出：[[1],[0,5],[2,3,4]]

 

提示：

• groupSizes.length == n
• 1 <= n <= 500
• 1 <= groupSizes[i] <= n

通过代码

高赞题解

class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        // key 用户组，value 为用户组 id
        Map<Integer, List<Integer>> map = new HashMap<>();
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < groupSizes.length; i++) {
            if (!map.containsKey(groupSizes[i])) {
                // 创建用户组
                map.put(groupSizes[i], new ArrayList<>());
            }
            // 将用户 id 放入对应的用户组
            List<Integer> sub = map.get(groupSizes[i]);
            sub.add(i);
            map.put(groupSizes[i], sub);
            // 当该用户组中的数量满了时，放入 result 集合，并清空 sub
            if (sub.size() == groupSizes[i]) {
                result.add(new ArrayList<>(sub));
                sub.clear();
            }
        }
        return result;
    }
}

统计信息

通过次数	提交次数	AC比率
12834	15828	81.1%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言