[LeetCode] 2610. Convert an Array Into a 2D Array With Conditions

You are given an integer array nums. You need to create a 2D array from nums satisfying the following conditions:

The 2D array should contain only the elements of the array nums.
Each row in the 2D array contains distinct integers.
The number of rows in the 2D array should be minimal.
Return the resulting array. If there are multiple answers, return any of them.

Note that the 2D array can have a different number of elements on each row.

Example 1:
Input: nums = [1,3,4,1,2,3,1]
Output: [[1,3,4,2],[1,3],[1]]
Explanation: We can create a 2D array that contains the following rows:

• 1,3,4,2
• 1,3
• 1
  All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
  It can be shown that we cannot have less than 3 rows in a valid array.

Example 2:
Input: nums = [1,2,3,4]
Output: [[4,3,2,1]]
Explanation: All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.

Constraints:
1 <= nums.length <= 200
1 <= nums[i] <= nums.length

转换二维数组。

给你一个整数数组 nums 。请你创建一个满足以下条件的二维数组：

二维数组应该 只 包含数组 nums 中的元素。
二维数组中的每一行都包含 不同 的整数。
二维数组的行数应尽可能 少 。
返回结果数组。如果存在多种答案，则返回其中任何一种。

请注意，二维数组的每一行上可以存在不同数量的元素。

思路

遍历 input 数组，用一个 hashmap 存每个不同数字的出现次数，同时用一个变量 row 记录最多的出现次数。这个 row 即为最终的二维数组的行数。

遍历 hashmap 里的每个元素，将出现次数为 i 的数字放到第 i 行，最后返回结果。

复杂度

时间O(n)
空间O(n)

代码

Java实现

class Solution {
    public List<List<Integer>> findMatrix(int[] nums) {
        int row = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
            row = Math.max(row, map.get(num));
        }
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < row; i++) {
            res.add(new ArrayList<>());
        }

        for (int num : map.keySet()) {
            int count = map.getOrDefault(num, 0);
            for (int j = 0; j < count; j++) {
                res.get(j).add(num);
            }
        }
        return res;
    }
}