> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/1282.group_the_people_given_the_group_size_they_belong_to.md).

# 1282.用户分组

## [To Index](https://github.com/newdee/Leetcode/blob/master/index.md)

## 1282. 用户分组

难度:Medium

> 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
```

```
class Solution {
public:
    vector<vector<int>> groupThePeople(vector<int>& groupSizes) {
        vector<vector<int>> res,tmp;
        tmp.resize(groupSizes.size());
        for(int i=0;i<groupSizes.size();i++)
        {
            tmp[groupSizes[i]-1].push_back(i);
        }
        for(int g =0; g<tmp.size();g++)
        {
            if(!tmp[g].empty())
            {
                for(int i=0;i<tmp[g].size()/(g+1);i++)
                {
                    vector<int>te;
                    for(int j=0;j<g+1;j++)
                    te.push_back(tmp[g][i*(g+1)+j]);
                    res.push_back(te);
                }

            }
        }
        return res;

    }
};
```

> 执行用时 :24 ms, 在所有 C++ 提交中击败了80.71%的用户\
> 内存消耗 :16.7 MB, 在所有 C++ 提交中击败了5.14%的用户


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