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feat: add solutions to lcof2 problem:No.075
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‎lcof2/剑指 Offer II 075. 数组相对排序/README.md

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@@ -42,22 +42,119 @@
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<!-- 这里可写通用的实现逻辑 -->
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由于本题数据范围是 `[0, 1000]`，因此我们可以开辟一个长度为 1001 的数组 mp，记录 arr1 的元素以及出现的次数。
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接着，遍历 arr2 中每个元素 x，若 `mp[x]` 大于 0，循环将 x 加入到答案数组 arr1 中，并且递减 `mp[x]`。遍历结束后，再从下标 `j = 0` 开始遍历 mp，若遇到 `mp[j]` 大于 0，将 `mp[j]` 个 j 加入到答案数组 arr1 中。
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最后返回 arr1 即可。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
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mp = {num: i for i, num in enumerate(arr2)}
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arr1.sort(key=lambda x: (mp.get(x, 10000), x))
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return arr1
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```
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```python
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class Solution:
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def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
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mp = [0] * 1001
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for x in arr1:
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mp[x] += 1
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i = 0
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for x in arr2:
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while mp[x] > 0:
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arr1[i] = x
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mp[x] -= 1
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i += 1
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for x, cnt in enumerate(mp):
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for _ in range(cnt):
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arr1[i] = x
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i += 1
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return arr1
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] relativeSortArray(int[] arr1, int[] arr2) {
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int[] mp = new int[1001];
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for (int x : arr1) {
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++mp[x];
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}
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int i = 0;
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for (int x : arr2) {
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while (mp[x]-- > 0) {
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arr1[i++] = x;
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}
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}
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for (int j = 0; j < mp.length; ++j) {
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while (mp[j]-- > 0) {
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arr1[i++] = j;
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}
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}
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return arr1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
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vector<int> mp(1001);
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for (int x : arr1) ++mp[x];
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int i = 0;
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for (int x : arr2)
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{
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while (mp[x]-- > 0) arr1[i++] = x;
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}
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for (int j = 0; j < mp.size(); ++j)
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{
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while (mp[j]-- > 0) arr1[i++] = j;
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}
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return arr1;
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}
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};
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```
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### **Go**
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```go
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func relativeSortArray(arr1 []int, arr2 []int) []int {
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mp := make([]int, 1001)
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for _, x := range arr1 {
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mp[x]++
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}
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i := 0
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for _, x := range arr2 {
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for mp[x] > 0 {
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arr1[i] = x
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mp[x]--
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i++
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}
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}
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for j, cnt := range mp {
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for cnt > 0 {
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arr1[i] = j
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i++
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cnt--
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}
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}
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return arr1
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}
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```
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### **...**

‎lcof2/剑指 Offer II 075. 数组相对排序/Solution.cpp

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class Solution {
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public:
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vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
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vector<int> mp(1001);
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for (int x : arr1) ++mp[x];
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int i = 0;
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for (int x : arr2)
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{
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while (mp[x]-- > 0) arr1[i++] = x;
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}
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for (int j = 0; j < mp.size(); ++j)
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{
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while (mp[j]-- > 0) arr1[i++] = j;
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}
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return arr1;
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}
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};

‎lcof2/剑指 Offer II 075. 数组相对排序/Solution.go

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func relativeSortArray(arr1 []int, arr2 []int) []int {
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mp := make([]int, 1001)
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for _, x := range arr1 {
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mp[x]++
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}
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i := 0
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for _, x := range arr2 {
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for mp[x] > 0 {
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arr1[i] = x
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mp[x]--
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i++
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}
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}
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for j, cnt := range mp {
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for cnt > 0 {
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arr1[i] = j
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i++
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cnt--
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}
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}
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return arr1
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}

‎lcof2/剑指 Offer II 075. 数组相对排序/Solution.java

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

‎lcof2/剑指 Offer II 075. 数组相对排序/Solution.py

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class Solution:
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def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
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mp = [0] * 1001
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for x in arr1:
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mp[x] += 1
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i = 0
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for x in arr2:
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while mp[x] > 0:
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arr1[i] = x
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mp[x] -= 1
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i += 1
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for x, cnt in enumerate(mp):
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for _ in range(cnt):
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arr1[i] = x
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i += 1
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return arr1

‎solution/1100-1199/1122.Relative Sort Array/README.md

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@@ -39,6 +39,12 @@
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<!-- 这里可写通用的实现逻辑 -->
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由于本题数据范围是 `[0, 1000]`，因此我们可以开辟一个长度为 1001 的数组 mp，记录 arr1 的元素以及出现的次数。
43+
44+
接着，遍历 arr2 中每个元素 x，若 `mp[x]` 大于 0，循环将 x 加入到答案数组 arr1 中，并且递减 `mp[x]`。遍历结束后，再从下标 `j = 0` 开始遍历 mp，若遇到 `mp[j]` 大于 0，将 `mp[j]` 个 j 加入到答案数组 arr1 中。
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最后返回 arr1 即可。
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<!-- tabs:start -->
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### **Python3**

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