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feat: add solutions to lc problem: No.0870
No.0870.Advantage Shuffle
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‎solution/0800-0899/0870.Advantage Shuffle/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 排序**
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类似田忌赛马。将 $nums1$, $nums2$ 按照升序排列。然后遍历 $nums1$ 中的每个元素 $v$，若在 $nums2[i..j]$ 中找不到比 $v$ 小的，则将 $v$ 与当前 $nums2[i..j]$ 中的最大元素匹配。
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时间复杂度 $O(nlogn)$。
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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 advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
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nums1.sort()
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t = [(v, i) for i, v in enumerate(nums2)]
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t.sort()
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n = len(nums2)
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ans = [0] * n
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i, j = 0, n - 1
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for v in nums1:
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if v <= t[i][0]:
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ans[t[j][1]] = v
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j -= 1
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else:
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ans[t[i][1]] = v
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i += 1
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return ans
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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[] advantageCount(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int[][] t = new int[n][2];
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for (int i = 0; i < n; ++i) {
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t[i] = new int[]{nums2[i], i};
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}
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Arrays.sort(t, (a, b) -> a[0] - b[0]);
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Arrays.sort(nums1);
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int[] ans = new int[n];
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int i = 0, j = n - 1;
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for (int v : nums1) {
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if (v <= t[i][0]) {
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ans[t[j--][1]] = v;
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} else {
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ans[t[i++][1]] = v;
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}
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}
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return ans;
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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> advantageCount(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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vector<pair<int, int>> t;
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for (int i = 0; i < n; ++i) t.push_back({nums2[i], i});
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sort(t.begin(), t.end());
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sort(nums1.begin(), nums1.end());
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int i = 0, j = n - 1;
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vector<int> ans(n);
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for (int v : nums1)
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{
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if (v <= t[i].first) ans[t[j--].second] = v;
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else ans[t[i++].second] = v;
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}
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return ans;
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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 advantageCount(nums1 []int, nums2 []int) []int {
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n := len(nums1)
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t := make([][]int, n)
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for i, v := range nums2 {
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t[i] = []int{v, i}
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}
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sort.Slice(t, func(i, j int) bool {
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return t[i][0] < t[j][0]
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})
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sort.Ints(nums1)
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ans := make([]int, n)
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i, j := 0, n-1
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for _, v := range nums1 {
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if v <= t[i][0] {
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ans[t[j][1]] = v
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j--
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} else {
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ans[t[i][1]] = v
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i++
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}
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}
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return ans
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}
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```
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### **...**

‎solution/0800-0899/0870.Advantage Shuffle/README_EN.md

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### **Python3**
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```python
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class Solution:
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def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
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nums1.sort()
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t = [(v, i) for i, v in enumerate(nums2)]
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t.sort()
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n = len(nums2)
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ans = [0] * n
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i, j = 0, n - 1
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for v in nums1:
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if v <= t[i][0]:
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ans[t[j][1]] = v
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j -= 1
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else:
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ans[t[i][1]] = v
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i += 1
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int[] advantageCount(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int[][] t = new int[n][2];
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for (int i = 0; i < n; ++i) {
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t[i] = new int[]{nums2[i], i};
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}
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Arrays.sort(t, (a, b) -> a[0] - b[0]);
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Arrays.sort(nums1);
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int[] ans = new int[n];
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int i = 0, j = n - 1;
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for (int v : nums1) {
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if (v <= t[i][0]) {
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ans[t[j--][1]] = v;
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} else {
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ans[t[i++][1]] = v;
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}
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}
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return ans;
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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> advantageCount(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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vector<pair<int, int>> t;
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for (int i = 0; i < n; ++i) t.push_back({nums2[i], i});
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sort(t.begin(), t.end());
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sort(nums1.begin(), nums1.end());
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int i = 0, j = n - 1;
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vector<int> ans(n);
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for (int v : nums1)
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{
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if (v <= t[i].first) ans[t[j--].second] = v;
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else ans[t[i++].second] = v;
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}
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return ans;
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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 advantageCount(nums1 []int, nums2 []int) []int {
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n := len(nums1)
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t := make([][]int, n)
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for i, v := range nums2 {
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t[i] = []int{v, i}
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}
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sort.Slice(t, func(i, j int) bool {
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return t[i][0] < t[j][0]
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})
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sort.Ints(nums1)
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ans := make([]int, n)
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i, j := 0, n-1
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for _, v := range nums1 {
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if v <= t[i][0] {
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ans[t[j][1]] = v
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j--
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} else {
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ans[t[i][1]] = v
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i++
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}
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}
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return ans
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}
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```
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### **...**

‎solution/0800-0899/0870.Advantage Shuffle/Solution.cpp

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class Solution {
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public:
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vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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vector<pair<int, int>> t;
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for (int i = 0; i < n; ++i) t.push_back({nums2[i], i});
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sort(t.begin(), t.end());
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sort(nums1.begin(), nums1.end());
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int i = 0, j = n - 1;
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vector<int> ans(n);
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for (int v : nums1)
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{
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if (v <= t[i].first) ans[t[j--].second] = v;
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else ans[t[i++].second] = v;
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}
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return ans;
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}
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};

‎solution/0800-0899/0870.Advantage Shuffle/Solution.go

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func advantageCount(nums1 []int, nums2 []int) []int {
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n := len(nums1)
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t := make([][]int, n)
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for i, v := range nums2 {
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t[i] = []int{v, i}
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}
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sort.Slice(t, func(i, j int) bool {
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return t[i][0] < t[j][0]
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})
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sort.Ints(nums1)
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ans := make([]int, n)
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i, j := 0, n-1
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for _, v := range nums1 {
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if v <= t[i][0] {
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ans[t[j][1]] = v
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j--
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} else {
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ans[t[i][1]] = v
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i++
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}
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}
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return ans
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}

‎solution/0800-0899/0870.Advantage Shuffle/Solution.java

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class Solution {
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public int[] advantageCount(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int[][] t = new int[n][2];
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for (int i = 0; i < n; ++i) {
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t[i] = new int[]{nums2[i], i};
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}
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Arrays.sort(t, (a, b) -> a[0] - b[0]);
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Arrays.sort(nums1);
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int[] ans = new int[n];
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int i = 0, j = n - 1;
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for (int v : nums1) {
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if (v <= t[i][0]) {
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ans[t[j--][1]] = v;
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} else {
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ans[t[i++][1]] = v;
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}
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}
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return ans;
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}
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}

‎solution/0800-0899/0870.Advantage Shuffle/Solution.py

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class Solution:
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def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
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nums1.sort()
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t = [(v, i) for i, v in enumerate(nums2)]
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t.sort()
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n = len(nums2)
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ans = [0] * n
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i, j = 0, n - 1
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for v in nums1:
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if v <= t[i][0]:
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ans[t[j][1]] = v
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j -= 1
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else:
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ans[t[i][1]] = v
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i += 1
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return ans

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