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feat: add solutions to lc problem: No.0982
No.0982.Triples with Bitwise AND Equal To Zero
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‎solution/0900-0999/0982.Triples with Bitwise AND Equal To Zero/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举 + 计数**
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我们可以先枚举任意两个数 $x$ 和 $y$，用哈希表或数组 $cnt$ 统计它们的按位与结果 $x \& y$ 出现的次数。
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然后我们枚举 $x$ 和 $y$ 的按位与结果 $xy$，再枚举 $z$，如果 $xy \& z = 0$，则将 $cnt[xy]$ 的值加入答案。
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最后返回答案即可。
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时间复杂度 $O(n^2 + n \times M)$，空间复杂度 $O(M)$，其中 $n$ 是数组 $nums$ 的长度；而 $M$ 是数组 $nums$ 中的最大值，本题中 $M \leq 2^{16}$。
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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 countTriplets(self, nums: List[int]) -> int:
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cnt = Counter(x & y for x in nums for y in nums)
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return sum(v for xy, v in cnt.items() for z in nums if xy & z == 0)
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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 countTriplets(int[] nums) {
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int mx = 0;
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for (int x : nums) {
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mx = Math.max(mx, x);
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}
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int[] cnt = new int[mx + 1];
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for (int x : nums) {
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for (int y : nums) {
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cnt[x & y]++;
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}
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}
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int ans = 0;
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for (int xy = 0; xy <= mx; ++xy) {
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for (int z : nums) {
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if ((xy & z) == 0) {
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ans += cnt[xy];
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}
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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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int countTriplets(vector<int>& nums) {
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int mx = *max_element(nums.begin(), nums.end());
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int cnt[mx + 1];
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memset(cnt, 0, sizeof cnt);
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for (int& x : nums) {
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for (int& y : nums) {
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cnt[x & y]++;
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}
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}
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int ans = 0;
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for (int xy = 0; xy <= mx; ++xy) {
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for (int& z : nums) {
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if ((xy & z) == 0) {
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ans += cnt[xy];
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}
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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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### **Go**
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```go
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func countTriplets(nums []int) (ans int) {
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mx := 0
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for _, x := range nums {
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mx = max(mx, x)
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}
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cnt := make([]int, mx+1)
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for _, x := range nums {
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for _, y := range nums {
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cnt[x&y]++
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}
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}
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for xy := 0; xy <= mx; xy++ {
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for _, z := range nums {
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if xy&z == 0 {
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ans += cnt[xy]
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}
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}
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0900-0999/0982.Triples with Bitwise AND Equal To Zero/README_EN.md

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### **Python3**
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```python
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class Solution:
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def countTriplets(self, nums: List[int]) -> int:
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cnt = Counter(x & y for x in nums for y in nums)
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return sum(v for xy, v in cnt.items() for z in nums if xy & z == 0)
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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 countTriplets(int[] nums) {
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int mx = 0;
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for (int x : nums) {
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mx = Math.max(mx, x);
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}
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int[] cnt = new int[mx + 1];
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for (int x : nums) {
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for (int y : nums) {
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cnt[x & y]++;
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}
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}
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int ans = 0;
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for (int xy = 0; xy <= mx; ++xy) {
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for (int z : nums) {
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if ((xy & z) == 0) {
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ans += cnt[xy];
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}
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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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int countTriplets(vector<int>& nums) {
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int mx = *max_element(nums.begin(), nums.end());
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int cnt[mx + 1];
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memset(cnt, 0, sizeof cnt);
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for (int& x : nums) {
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for (int& y : nums) {
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cnt[x & y]++;
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}
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}
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int ans = 0;
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for (int xy = 0; xy <= mx; ++xy) {
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for (int& z : nums) {
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if ((xy & z) == 0) {
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ans += cnt[xy];
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}
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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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### **Go**
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```go
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func countTriplets(nums []int) (ans int) {
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mx := 0
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for _, x := range nums {
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mx = max(mx, x)
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}
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cnt := make([]int, mx+1)
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for _, x := range nums {
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for _, y := range nums {
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cnt[x&y]++
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}
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}
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for xy := 0; xy <= mx; xy++ {
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for _, z := range nums {
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if xy&z == 0 {
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ans += cnt[xy]
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}
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}
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0900-0999/0982.Triples with Bitwise AND Equal To Zero/Solution.cpp

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class Solution {
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public:
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int countTriplets(vector<int>& nums) {
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int mx = *max_element(nums.begin(), nums.end());
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int cnt[mx + 1];
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memset(cnt, 0, sizeof cnt);
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for (int& x : nums) {
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for (int& y : nums) {
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cnt[x & y]++;
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}
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}
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int ans = 0;
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for (int xy = 0; xy <= mx; ++xy) {
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for (int& z : nums) {
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if ((xy & z) == 0) {
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ans += cnt[xy];
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}
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}
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}
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return ans;
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}
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};

‎solution/0900-0999/0982.Triples with Bitwise AND Equal To Zero/Solution.go

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func countTriplets(nums []int) (ans int) {
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mx := 0
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for _, x := range nums {
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mx = max(mx, x)
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}
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cnt := make([]int, mx+1)
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for _, x := range nums {
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for _, y := range nums {
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cnt[x&y]++
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}
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}
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for xy := 0; xy <= mx; xy++ {
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for _, z := range nums {
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if xy&z == 0 {
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ans += cnt[xy]
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}
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}
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/0900-0999/0982.Triples with Bitwise AND Equal To Zero/Solution.java

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class Solution {
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public int countTriplets(int[] nums) {
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int mx = 0;
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for (int x : nums) {
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mx = Math.max(mx, x);
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}
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int[] cnt = new int[mx + 1];
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for (int x : nums) {
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for (int y : nums) {
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cnt[x & y]++;
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}
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}
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int ans = 0;
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for (int xy = 0; xy <= mx; ++xy) {
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for (int z : nums) {
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if ((xy & z) == 0) {
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ans += cnt[xy];
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}
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}
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}
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return ans;
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}
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}

‎solution/0900-0999/0982.Triples with Bitwise AND Equal To Zero/Solution.py

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class Solution:
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def countTriplets(self, nums: List[int]) -> int:
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cnt = Counter(x & y for x in nums for y in nums)
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return sum(v for xy, v in cnt.items() for z in nums if xy & z == 0)

‎solution/1800-1899/1836.Remove Duplicates From an Unsorted Linked List/README_EN.md

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<p><strong>Constraints:</strong></p>
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<ul>
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<li>The number of nodes in the list is in the range&nbsp;<code>[1, 10<sup>5</sup>]</code></li>
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<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
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</ul>
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## Solutions

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