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# 1122.Relative-Sort-Array

## 题目描述

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.

## 代码

### Approach #1 Replace elements according to the count

Time: O(1) && Space: O(1)
class Solution {
public int[] relativeSortArray(int[] arr1, int[] arr2) {
int[] cnt = new int;
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;
}
}

### Approach #2 TreeMap

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) {
while (map.get(n) > 0) {
arr1[i++] = n;
map.put(n, map.get(n) - 1);
}
}
for (int n: map.keySet()) {
while (map.get(n) > 0) {
arr1[i++] = n;
map.put(n, map.get(n) - 1);
}
}
return arr1;
}
}