数组：

给定两个数组，arr1 和 arr2，

arr2 中的元素各不相同
arr2 中的每个元素都出现在 arr1 中
对 arr1 中的元素进行排序，使 arr1 中项的相对顺序和 arr2 中的相对顺序相同。未在 arr2 中出现过的元素需要按照升序放在 arr1 的末尾

示例：

输入：arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
输出：[2,2,2,1,4,3,3,9,6,7,19]

提示：

1 <= arr1.length, arr2.length <= 1000
0 <= arr1[i], arr2[i] <= 1000
arr2 中的元素 arr2[i] 各不相同
arr2 中的每个元素 arr2[i] 都出现在 arr1 中

public class Solution {
    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] arr1 = new int[]{1, 3, 2, 8, 10, 4, 6, 19, 9, 11, 7};
        int[] arr2 = new int[]{2, 1, 4, 3, 9, 6};
        System.out.println(Arrays.stream(solution.relativeSortArray(arr1, arr2)).boxed().collect(Collectors.toList()));
    }

    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        int left = 0;//需要替换的索引
        int right;//arr1需要判断的索引
        int swap;//交换的值
        for (int i = 0; i < arr2.length; i++) {
            right = 0;
            while (right < arr1.length) {
                if (arr1[right] == arr2[i]) {//相等
                    swap = arr1[left];
                    arr1[left] = arr1[right];
                    arr1[right] = swap;
                    left++;
                }
                right++;
            }
        }
        for (int i = left; i < arr1.length-1; i++) {
            left = i;//前面已经根据arr2排序完了，根据升序排序剩余的
            right = i+1;
            while (right < arr1.length){
                if(arr1[right]<arr1[left]){
                    swap = arr1[left];
                    arr1[left]=arr1[right];
                    arr1[right] = swap;
                }
                right++;
            }
        }
        return arr1;
    }
}

题目出自LeetCode