算法: 获取两个数组的公共数字1122. Relative Sort Array

Posted 2021-12-06 架构师易筋

1122. 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 do not 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]

Example 2:

Input: arr1 = [28,6,22,8,44,17], arr2 = [22,28,8,6]
Output: [22,28,8,6,17,44]

Constraints:

• 1 <= arr1.length, arr2.length <= 1000
• 0 <= arr1[i], arr2[i] <= 1000
• All the elements of arr2 are distinct.
• Each arr2[i] is in arr1.

利用数字范围[0, 1000]简化解法

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        int[] cnt = new int[1001];
        int i = 0;
        for (int a1: arr1) cnt[a1]++;
        for (int a2: arr2) {
            while (cnt[a2]-- > 0) arr1[i++] = a2;
        }
        for (int k = 0; k < 1001; k++) {
            while (cnt[k]-- > 0) arr1[i++] = k;
        }
            
        return arr1;
    }
}

如果没有数字范围的约束

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for(int n : arr1) map.put(n, map.getOrDefault(n, 0) + 1);
        int i = 0;
        for(int n : arr2) {
            for(int j = 0; j < map.get(n); j++) {
                arr1[i++] = n;
            }
            map.remove(n);
        }
        for(int n : map.keySet()){
            for(int j = 0; j < map.get(n); j++) {
                arr1[i++] = n;
            }
        }
        return arr1;
    }
}

参考

https://leetcode.com/problems/relative-sort-array/discuss/335056/Java-in-place-solution-using-counting-sort