描述
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.
解析
根据题意将结果分为两部分，左部分是在 arr1 和 arr2 中都包含的元素，右部分是 arr1 中存在但是 arr2 中不存在的元素，遍历寻找就好，时间复杂度为 O(M * N)，M 和 N 分别是 两个数组的长度,空间复杂度为 O(N)。
解答
class Solution(object):
def relativeSortArray(self, arr1, arr2):
"""
:type arr1: List[int]
:type arr2: List[int]
:rtype: List[int]
"""
left = []
right = []
counter = collections.Counter(arr1)
for c in arr2:
if c in arr1 :
left += [c]*counter[c]
for c in arr1:
if c not in arr2 and c not in right:
right +=[c]*counter[c]
right.sort()
return left+right
运行结果
Runtime: 28 ms, faster than 41.96% of Python online submissions for Relative Sort Array.
Memory Usage: 12 MB, less than 100.00% of Python online submissions for Relative Sort Array.
