Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

feat: add solutions to lc problem: No.0532 #4164

Merged
merged 1 commit into from
Mar 15, 2025
Merged

feat: add solutions to lc problem: No.0532 #4164

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
137 changes: 70 additions & 67 deletions solution/0500-0599/0532.K-diff Pairs in an Array/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -77,11 +77,13 @@ tags:

### 方法一：哈希表

由于 $k$ 是一个定值，因此用哈希表 $ans$ 记录数对的较小值，就能够确定较大的值。最后返回 ans 的大小作为答案。
由于 $k$ 是一个定值，我们可以用一个哈希表 $\textit{ans}$ 记录数对的较小值，就能够确定较大的值。最后返回 $\textit{ans}$ 的大小作为答案。

遍历数组 $nums$，当前遍历到的数 $nums[j]$，我们记为 $v$，用哈希表 $vis$ 记录此前遍历到的所有数字。若 $v-k$ 在 $vis$ 中，则将 $v-k$ 添加至 $ans$；若 $v+k$ 在 $vis$ 中，则将 $v$ 添加至 $ans$。
遍历数组 $\textit{nums}$，当前遍历到的数 $x$，我们用哈希表 $\textit{vis}$ 记录此前遍历到的所有数字。若 $x-k$ 在 $\textit{vis}$ 中，则将 $x-k$ 添加至 $\textit{ans}$；若 $x+k$ 在 $\textit{vis}$ 中，则将 $x$ 添加至 $\textit{ans}$。然后我们将 $x$ 添加至 $\textit{vis}$。继续遍历数组 $\textit{nums}$ 直至遍历结束。

时间复杂度 $O(n)$，其中 $n$ 表示数组 $nums$ 的长度。
最后返回 $\textit{ans}$ 的大小作为答案。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。

<!-- tabs:start -->

Expand All @@ -90,13 +92,14 @@ tags:
```python
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
vis, ans = set(), set()
for v in nums:
if v - k in vis:
ans.add(v - k)
if v + k in vis:
ans.add(v)
vis.add(v)
ans = set()
vis = set()
for x in nums:
if x - k in vis:
ans.add(x - k)
if x + k in vis:
ans.add(x)
vis.add(x)
return len(ans)
```

Expand All @@ -105,16 +108,16 @@ class Solution:
```java
class Solution {
public int findPairs(int[] nums, int k) {
Set<Integer> vis = new HashSet<>();
Set<Integer> ans = new HashSet<>();
for (int v : nums) {
if (vis.contains(v - k)) {
ans.add(v - k);
Set<Integer> vis = new HashSet<>();
for (int x : nums) {
if (vis.contains(x - k)) {
ans.add(x - k);
}
if (vis.contains(v + k)) {
ans.add(v);
if (vis.contains(x + k)) {
ans.add(x);
}
vis.add(v);
vis.add(x);
}
return ans.size();
}
Expand All @@ -127,12 +130,15 @@ class Solution {
class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_set<int> vis;
unordered_set<int> ans;
for (int& v : nums) {
if (vis.count(v - k)) ans.insert(v - k);
if (vis.count(v + k)) ans.insert(v);
vis.insert(v);
unordered_set<int> ans, vis;
for (int x : nums) {
if (vis.count(x - k)) {
ans.insert(x - k);
}
if (vis.count(x + k)) {
ans.insert(x);
}
vis.insert(x);
}
return ans.size();
}
Expand All @@ -143,52 +149,61 @@ public:

```go
func findPairs(nums []int, k int) int {
vis := map[int]bool{}
ans := map[int]bool{}
for _, v := range nums {
if vis[v-k] {
ans[v-k] = true
ans := make(map[int]struct{})
vis := make(map[int]struct{})

for _, x := range nums {
if _, ok := vis[x-k]; ok {
ans[x-k] = struct{}{}
}
if vis[v+k] {
ans[v] = true
if _, ok := vis[x+k]; ok {
ans[x] = struct{}{}
}
vis[v] = true
vis[x] = struct{}{}
}
return len(ans)
}
```

#### TypeScript

```ts
function findPairs(nums: number[], k: number): number {
const ans = new Set<number>();
const vis = new Set<number>();
for (const x of nums) {
if (vis.has(x - k)) {
ans.add(x - k);
}
if (vis.has(x + k)) {
ans.add(x);
}
vis.add(x);
}
return ans.size;
}
```

#### Rust

```rust
use std::collections::HashSet;

impl Solution {
pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
nums.sort();
let n = nums.len();
let mut res = 0;
let mut left = 0;
let mut right = 1;
while right < n {
let num = i32::abs(nums[left] - nums[right]);
if num == k {
res += 1;
pub fn find_pairs(nums: Vec<i32>, k: i32) -> i32 {
let mut ans = HashSet::new();
let mut vis = HashSet::new();

for &x in &nums {
if vis.contains(&(x - k)) {
ans.insert(x - k);
}
if num <= k {
right += 1;
while right < n && nums[right - 1] == nums[right] {
right += 1;
}
} else {
left += 1;
while left < right && nums[left - 1] == nums[left] {
left += 1;
}
if left == right {
right += 1;
}
if vis.contains(&(x + k)) {
ans.insert(x);
}
vis.insert(x);
}
res
ans.len() as i32
}
}
```
Expand All @@ -197,16 +212,4 @@ impl Solution {

<!-- solution:end -->

<!-- solution:start -->

### 方法二：排序 + 双指针

只需要统计组合的数量，因此可以改动原数组，对其排序，使用双指针来统计。

声明 `left` 与 `right` 指针，初始化为 0 和 1。根据 `abs(nums[left] - nums[right])` 与 `k` 值对比结果移动指针。

需要注意的是，**不能出现重复的组合**，所以移动指针时，不能仅仅是 `+1`，需要到一个不等于当前值的位置。

<!-- solution:end -->

<!-- problem:end -->
127 changes: 74 additions & 53 deletions solution/0500-0599/0532.K-diff Pairs in an Array/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -73,7 +73,15 @@ Although we have two 1s in the input, we should only return the number of <stron

<!-- solution:start -->

### Solution 1
### Solution 1: Hash Table

Since $k$ is a fixed value, we can use a hash table $\textit{ans}$ to record the smaller value of the pairs, which allows us to determine the larger value. Finally, we return the size of $\textit{ans}$ as the answer.

We traverse the array $\textit{nums}$. For the current number $x$, we use a hash table $\textit{vis}$ to record all the numbers that have been traversed. If $x-k$ is in $\textit{vis}$, we add $x-k$ to $\textit{ans}$. If $x+k$ is in $\textit{vis}$, we add $x$ to $\textit{ans}$. Then, we add $x$ to $\textit{vis}$. Continue traversing the array $\textit{nums}$ until the end.

Finally, we return the size of $\textit{ans}$ as the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

<!-- tabs:start -->

Expand All @@ -82,13 +90,14 @@ Although we have two 1s in the input, we should only return the number of <stron
```python
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
vis, ans = set(), set()
for v in nums:
if v - k in vis:
ans.add(v - k)
if v + k in vis:
ans.add(v)
vis.add(v)
ans = set()
vis = set()
for x in nums:
if x - k in vis:
ans.add(x - k)
if x + k in vis:
ans.add(x)
vis.add(x)
return len(ans)
```

Expand All @@ -97,16 +106,16 @@ class Solution:
```java
class Solution {
public int findPairs(int[] nums, int k) {
Set<Integer> vis = new HashSet<>();
Set<Integer> ans = new HashSet<>();
for (int v : nums) {
if (vis.contains(v - k)) {
ans.add(v - k);
Set<Integer> vis = new HashSet<>();
for (int x : nums) {
if (vis.contains(x - k)) {
ans.add(x - k);
}
if (vis.contains(v + k)) {
ans.add(v);
if (vis.contains(x + k)) {
ans.add(x);
}
vis.add(v);
vis.add(x);
}
return ans.size();
}
Expand All @@ -119,12 +128,15 @@ class Solution {
class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_set<int> vis;
unordered_set<int> ans;
for (int& v : nums) {
if (vis.count(v - k)) ans.insert(v - k);
if (vis.count(v + k)) ans.insert(v);
vis.insert(v);
unordered_set<int> ans, vis;
for (int x : nums) {
if (vis.count(x - k)) {
ans.insert(x - k);
}
if (vis.count(x + k)) {
ans.insert(x);
}
vis.insert(x);
}
return ans.size();
}
Expand All @@ -135,52 +147,61 @@ public:

```go
func findPairs(nums []int, k int) int {
vis := map[int]bool{}
ans := map[int]bool{}
for _, v := range nums {
if vis[v-k] {
ans[v-k] = true
ans := make(map[int]struct{})
vis := make(map[int]struct{})

for _, x := range nums {
if _, ok := vis[x-k]; ok {
ans[x-k] = struct{}{}
}
if vis[v+k] {
ans[v] = true
if _, ok := vis[x+k]; ok {
ans[x] = struct{}{}
}
vis[v] = true
vis[x] = struct{}{}
}
return len(ans)
}
```

#### TypeScript

```ts
function findPairs(nums: number[], k: number): number {
const ans = new Set<number>();
const vis = new Set<number>();
for (const x of nums) {
if (vis.has(x - k)) {
ans.add(x - k);
}
if (vis.has(x + k)) {
ans.add(x);
}
vis.add(x);
}
return ans.size;
}
```

#### Rust

```rust
use std::collections::HashSet;

impl Solution {
pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
nums.sort();
let n = nums.len();
let mut res = 0;
let mut left = 0;
let mut right = 1;
while right < n {
let num = i32::abs(nums[left] - nums[right]);
if num == k {
res += 1;
pub fn find_pairs(nums: Vec<i32>, k: i32) -> i32 {
let mut ans = HashSet::new();
let mut vis = HashSet::new();

for &x in &nums {
if vis.contains(&(x - k)) {
ans.insert(x - k);
}
if num <= k {
right += 1;
while right < n && nums[right - 1] == nums[right] {
right += 1;
}
} else {
left += 1;
while left < right && nums[left - 1] == nums[left] {
left += 1;
}
if left == right {
right += 1;
}
if vis.contains(&(x + k)) {
ans.insert(x);
}
vis.insert(x);
}
res
ans.len() as i32
}
}
```
Expand Down
17 changes: 10 additions & 7 deletions solution/0500-0599/0532.K-diff Pairs in an Array/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,16 @@
class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_set<int> vis;
unordered_set<int> ans;
for (int& v : nums) {
if (vis.count(v - k)) ans.insert(v - k);
if (vis.count(v + k)) ans.insert(v);
vis.insert(v);
unordered_set<int> ans, vis;
for (int x : nums) {
if (vis.count(x - k)) {
ans.insert(x - k);
}
if (vis.count(x + k)) {
ans.insert(x);
}
vis.insert(x);
}
return ans.size();
}
};
};
Loading