关键词

计数排序，指定顺序排序，TreeMap,countsort
计数排序讲解 https://www.cnblogs.com/kyoner/p/10604781.html

题目描述

https://leetcode.com/problems/relative-sort-array/

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]
 

Constraints:

arr1.length, arr2.length <= 1000
0 <= arr1[i], arr2[i] <= 1000
Each arr2[i] is distinct.
Each arr2[i] is in arr1.

博主提交的代码

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        Map<Integer,Integer> cache = new HashMap<>();
        for(int each: arr2){
            cache.put(each,0);
        }
        for(int i = 0; i < arr1.length; i++){
            if( cache.get(arr1[i]) != null){
                cache.put(arr1[i],cache.get(arr1[i]) + 1);
                arr1[i] = -1;
            }
        }
        Arrays.sort(arr1);
        
        int index = 0;
        for(int each: arr2){
            int count  = cache.get(each);
            while(count > 0){
                arr1[index] = each;
                index++;
                count--;
            }
        }
        return arr1;
    }
}

其他人优秀的代码

解法一

计数排序，这东西真的费内存，但是运行时间真的牛逼
https://leetcode.com/problems/relative-sort-array/discuss/335056/Java-in-place-solution-using-counting-sort

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        int[] cnt = new int[1001];
        for(int n : arr1) cnt[n]++;
        int i = 0;
        for(int n : arr2) {
            while(cnt[n]-- > 0) {
                arr1[i++] = n;
            }
        }
        for(int n = 0; n < cnt.length; n++) {
            while(cnt[n]-- > 0) {
                arr1[i++] = n;
            }
        }
        return arr1;
    }
}

解法二

https://leetcode.com/problems/relative-sort-array/discuss/335217/Java-O(n*lgn)-1-ms-beats-100

public int[] relativeSortArray(int[] arr1, int[] arr2) {
        //create map for counting numbers in arr1. Initialize everything with zeroes
        Map<Integer, Integer> m = new HashMap();
        for (int num : arr2) {
            m.put(num, 0);
        }
        int last = arr1.length - 1;
        int[] res = new int[arr1.length];
        //iterate over arr1  and count numbers of time this element is in arr1
        for (int num : arr1) {
            //if number is from arr2 - increment count
            if (m.containsKey(num))
                m.put(num, m.get(num) + 1);
            //otherwise add element to the end of res and decrement the pointer
            else {
                res[last--] = num;
            }
        }
        //iterate over arr2, fill elements in res based on it's count 
        int p = 0;
        for (int num : arr2) {
            int c = m.get(num);
            for (int i = 0; i < c; i++) {
                res[p++] = num;
            }
        }
        //now sort the leftovers - p points to the first element in series of those from arr2 that are not in arr1 
        Arrays.sort(res, p, res.length);
        return res;
    }

更加极致

public static int[] relativeSortArrayV1(int[] arr1, int[] arr2) {
        TreeMap<Integer, Integer> freq = new TreeMap<>();
        for (int num : arr1) {
            freq.merge(num, 1, (a, b) -> a + 1);
        }
        int[] res = new int[arr1.length];
        int i = 0;
        for (int key : arr2) {
            int reps = freq.remove(key);
            for (int j = 0; j < reps; j++) {
                res[i++] = key;
            }
        }

        while (freq.size() > 0) {
            int key = freq.firstKey();
            int reps = freq.remove(key);
            for (int j = 0; j < reps; j++) {
                res[i++] = key;
            }
        }
        return res;
    }