1122. Relative Sort Array

Description

Given two arrays arr1 and arr2, the elements of arr2 are distinct, and all elements in arr2 are also in arr1.

Sort the elements of arr1 such that the relative ordering of items in arr1 are the same as in arr2. Elements that don't appear in arr2 should be placed at the end of arr1 in ascending order

Example 1:

Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
Output: [2,2,2,1,4,3,3,9,6,7,19]

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Note

Links

(en)https://leetcode.com/problems/relative-sort-array/
(中文)https://leetcode-cn.com/problems/relative-sort-array/

Solutions

Solution1

使用一个 Map 统计 arr2 出现的次数，然后遍历 arr1，进行统计，把剩余的元素使用一个 List 进行存放并进行排序（Collections.sort）。

最后将这两部分进行整合即可。

Java Code

class Solution {
    /**
    approach one:
    Complexity Analysis: 
        Time complexity: O(n log n)
        Spacce complexity: O(n)
    */
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> list = new ArrayList<>();
        
        // put elemts into the map for counting
        for (int n : arr2) {
            map.put(n , 0);
        }
        for (int n : arr1) {
            if (map.containsKey(n)) {
                map.put(n, map.get(n) + 1);
            } else {
                list.add(n);
            }
        }
        int[] result = new int[arr1.length];
        int index = 0;
        // combine them together
        for (int n : arr2) {
            for (int i = 0; i < map.get(n); i++) {
                result[index++] = n;
            }
        }
        // sort the rest of elements. using mergeSort built-in in Collections tool.
        Collections.sort(list, (o1, o2) -> o1.compareTo(o2));
        for (int n : list) {
            result[index++] = n;
        }
        return result;
    }
}

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Submission status

solution-1-status

可以看到这一解决方式在速度方面是非常不好的，虽然可以解决问题。

# 1122. Relative Sort Array

# Description

# Solutions

# Solution1