1282. Group the People Given the Group Size They Belong To

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題目敘述

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]]

解題思路

• Map.computeIfAbsent(key, k -> value)
• Map.values()
• List.subList

Solution

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        Map<Integer, List<Integer>> groupMap = new HashMap<>();
        List<List<Integer>> result = new ArrayList<>();

        for (int idx = 0; idx < groupSizes.length; idx++) {
            int size = groupSizes[idx];
            // If groupMap didn't have the key of size then create new ArrayList with this size.
            groupMap.computeIfAbsent(size, k -> new ArrayList<>()).add(idx);
        }

        for (List<Integer> group : groupMap.values()) {
            int size = group.size();
            
            for (int i = 0; i < size; i += groupSizes[group.get(i)]) {
                result.add(group.subList(i, i + groupSizes[group.get(i)]));
            }
        }

        return result;
    }
}