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feat: add solutions to lc problem: No.0532 (doocs#4164)
No.0532.K-diff Pairs in an Array
1 parent 1b22bc7 commit 66f296d

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‎solution/0500-0599/0532.K-diff Pairs in an Array/README.md

Lines changed: 70 additions & 67 deletions
Original file line numberDiff line numberDiff line change
@@ -77,11 +77,13 @@ tags:
7777

7878
### 方法一：哈希表
7979

80-
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。
80+
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。
8181

82-
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 $ans$。
82+
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。
8383

84-
时间复杂度 $O(n)$，其中 $n$ 表示数组 $nums$ 的长度。
84+
最后返回 $\textit{ans}$ 的大小作为答案。
85+
86+
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。
8587

8688
<!-- tabs:start -->
8789

@@ -90,13 +92,14 @@ tags:
9092
```python
9193
class Solution:
9294
def findPairs(self, nums: List[int], k: int) -> int:
93-
vis, ans = set(), set()
94-
for v in nums:
95-
if v - k in vis:
96-
ans.add(v - k)
97-
if v + k in vis:
98-
ans.add(v)
99-
vis.add(v)
95+
ans = set()
96+
vis = set()
97+
for x in nums:
98+
if x - k in vis:
99+
ans.add(x - k)
100+
if x + k in vis:
101+
ans.add(x)
102+
vis.add(x)
100103
return len(ans)
101104
```
102105

@@ -105,16 +108,16 @@ class Solution:
105108
```java
106109
class Solution {
107110
public int findPairs(int[] nums, int k) {
108-
Set<Integer> vis = new HashSet<>();
109111
Set<Integer> ans = new HashSet<>();
110-
for (int v : nums) {
111-
if (vis.contains(v - k)) {
112-
ans.add(v - k);
112+
Set<Integer> vis = new HashSet<>();
113+
for (int x : nums) {
114+
if (vis.contains(x - k)) {
115+
ans.add(x - k);
113116
}
114-
if (vis.contains(v + k)) {
115-
ans.add(v);
117+
if (vis.contains(x + k)) {
118+
ans.add(x);
116119
}
117-
vis.add(v);
120+
vis.add(x);
118121
}
119122
return ans.size();
120123
}
@@ -127,12 +130,15 @@ class Solution {
127130
class Solution {
128131
public:
129132
int findPairs(vector<int>& nums, int k) {
130-
unordered_set<int> vis;
131-
unordered_set<int> ans;
132-
for (int& v : nums) {
133-
if (vis.count(v - k)) ans.insert(v - k);
134-
if (vis.count(v + k)) ans.insert(v);
135-
vis.insert(v);
133+
unordered_set<int> ans, vis;
134+
for (int x : nums) {
135+
if (vis.count(x - k)) {
136+
ans.insert(x - k);
137+
}
138+
if (vis.count(x + k)) {
139+
ans.insert(x);
140+
}
141+
vis.insert(x);
136142
}
137143
return ans.size();
138144
}
@@ -143,52 +149,61 @@ public:
143149
144150
```go
145151
func findPairs(nums []int, k int) int {
146-
vis := map[int]bool{}
147-
ans := map[int]bool{}
148-
for _, v := range nums {
149-
if vis[v-k] {
150-
ans[v-k] = true
152+
ans := make(map[int]struct{})
153+
vis := make(map[int]struct{})
154+
155+
for _, x := range nums {
156+
if _, ok := vis[x-k]; ok {
157+
ans[x-k] = struct{}{}
151158
}
152-
if vis[v+k] {
153-
ans[v] = true
159+
if _, ok := vis[x+k]; ok {
160+
ans[x] = struct{}{}
154161
}
155-
vis[v] = true
162+
vis[x] = struct{}{}
156163
}
157164
return len(ans)
158165
}
159166
```
160167

168+
#### TypeScript
169+
170+
```ts
171+
function findPairs(nums: number[], k: number): number {
172+
const ans = new Set<number>();
173+
const vis = new Set<number>();
174+
for (const x of nums) {
175+
if (vis.has(x - k)) {
176+
ans.add(x - k);
177+
}
178+
if (vis.has(x + k)) {
179+
ans.add(x);
180+
}
181+
vis.add(x);
182+
}
183+
return ans.size;
184+
}
185+
```
186+
161187
#### Rust
162188

163189
```rust
190+
use std::collections::HashSet;
191+
164192
impl Solution {
165-
pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
166-
nums.sort();
167-
let n = nums.len();
168-
let mut res = 0;
169-
let mut left = 0;
170-
let mut right = 1;
171-
while right < n {
172-
let num = i32::abs(nums[left] - nums[right]);
173-
if num == k {
174-
res += 1;
193+
pub fn find_pairs(nums: Vec<i32>, k: i32) -> i32 {
194+
let mut ans = HashSet::new();
195+
let mut vis = HashSet::new();
196+
197+
for &x in &nums {
198+
if vis.contains(&(x - k)) {
199+
ans.insert(x - k);
175200
}
176-
if num <= k {
177-
right += 1;
178-
while right < n && nums[right - 1] == nums[right] {
179-
right += 1;
180-
}
181-
} else {
182-
left += 1;
183-
while left < right && nums[left - 1] == nums[left] {
184-
left += 1;
185-
}
186-
if left == right {
187-
right += 1;
188-
}
201+
if vis.contains(&(x + k)) {
202+
ans.insert(x);
189203
}
204+
vis.insert(x);
190205
}
191-
res
206+
ans.len() as i32
192207
}
193208
}
194209
```
@@ -197,16 +212,4 @@ impl Solution {
197212

198213
<!-- solution:end -->
199214

200-
<!-- solution:start -->
201-
202-
### 方法二：排序 + 双指针
203-
204-
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。
205-
206-
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。
207-
208-
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。
209-
210-
<!-- solution:end -->
211-
212215
<!-- problem:end -->

‎solution/0500-0599/0532.K-diff Pairs in an Array/README_EN.md

Lines changed: 74 additions & 53 deletions
Original file line numberDiff line numberDiff line change
@@ -73,7 +73,15 @@ Although we have two 1s in the input, we should only return the number of <stron
7373

7474
<!-- solution:start -->
7575

76-
### Solution 1
76+
### Solution 1: Hash Table
77+
78+
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.
79+
80+
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.
81+
82+
Finally, we return the size of $\textit{ans}$ as the answer.
83+
84+
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
7785

7886
<!-- tabs:start -->
7987

@@ -82,13 +90,14 @@ Although we have two 1s in the input, we should only return the number of <stron
8290
```python
8391
class Solution:
8492
def findPairs(self, nums: List[int], k: int) -> int:
85-
vis, ans = set(), set()
86-
for v in nums:
87-
if v - k in vis:
88-
ans.add(v - k)
89-
if v + k in vis:
90-
ans.add(v)
91-
vis.add(v)
93+
ans = set()
94+
vis = set()
95+
for x in nums:
96+
if x - k in vis:
97+
ans.add(x - k)
98+
if x + k in vis:
99+
ans.add(x)
100+
vis.add(x)
92101
return len(ans)
93102
```
94103

@@ -97,16 +106,16 @@ class Solution:
97106
```java
98107
class Solution {
99108
public int findPairs(int[] nums, int k) {
100-
Set<Integer> vis = new HashSet<>();
101109
Set<Integer> ans = new HashSet<>();
102-
for (int v : nums) {
103-
if (vis.contains(v - k)) {
104-
ans.add(v - k);
110+
Set<Integer> vis = new HashSet<>();
111+
for (int x : nums) {
112+
if (vis.contains(x - k)) {
113+
ans.add(x - k);
105114
}
106-
if (vis.contains(v + k)) {
107-
ans.add(v);
115+
if (vis.contains(x + k)) {
116+
ans.add(x);
108117
}
109-
vis.add(v);
118+
vis.add(x);
110119
}
111120
return ans.size();
112121
}
@@ -119,12 +128,15 @@ class Solution {
119128
class Solution {
120129
public:
121130
int findPairs(vector<int>& nums, int k) {
122-
unordered_set<int> vis;
123-
unordered_set<int> ans;
124-
for (int& v : nums) {
125-
if (vis.count(v - k)) ans.insert(v - k);
126-
if (vis.count(v + k)) ans.insert(v);
127-
vis.insert(v);
131+
unordered_set<int> ans, vis;
132+
for (int x : nums) {
133+
if (vis.count(x - k)) {
134+
ans.insert(x - k);
135+
}
136+
if (vis.count(x + k)) {
137+
ans.insert(x);
138+
}
139+
vis.insert(x);
128140
}
129141
return ans.size();
130142
}
@@ -135,52 +147,61 @@ public:
135147
136148
```go
137149
func findPairs(nums []int, k int) int {
138-
vis := map[int]bool{}
139-
ans := map[int]bool{}
140-
for _, v := range nums {
141-
if vis[v-k] {
142-
ans[v-k] = true
150+
ans := make(map[int]struct{})
151+
vis := make(map[int]struct{})
152+
153+
for _, x := range nums {
154+
if _, ok := vis[x-k]; ok {
155+
ans[x-k] = struct{}{}
143156
}
144-
if vis[v+k] {
145-
ans[v] = true
157+
if _, ok := vis[x+k]; ok {
158+
ans[x] = struct{}{}
146159
}
147-
vis[v] = true
160+
vis[x] = struct{}{}
148161
}
149162
return len(ans)
150163
}
151164
```
152165

166+
#### TypeScript
167+
168+
```ts
169+
function findPairs(nums: number[], k: number): number {
170+
const ans = new Set<number>();
171+
const vis = new Set<number>();
172+
for (const x of nums) {
173+
if (vis.has(x - k)) {
174+
ans.add(x - k);
175+
}
176+
if (vis.has(x + k)) {
177+
ans.add(x);
178+
}
179+
vis.add(x);
180+
}
181+
return ans.size;
182+
}
183+
```
184+
153185
#### Rust
154186

155187
```rust
188+
use std::collections::HashSet;
189+
156190
impl Solution {
157-
pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
158-
nums.sort();
159-
let n = nums.len();
160-
let mut res = 0;
161-
let mut left = 0;
162-
let mut right = 1;
163-
while right < n {
164-
let num = i32::abs(nums[left] - nums[right]);
165-
if num == k {
166-
res += 1;
191+
pub fn find_pairs(nums: Vec<i32>, k: i32) -> i32 {
192+
let mut ans = HashSet::new();
193+
let mut vis = HashSet::new();
194+
195+
for &x in &nums {
196+
if vis.contains(&(x - k)) {
197+
ans.insert(x - k);
167198
}
168-
if num <= k {
169-
right += 1;
170-
while right < n && nums[right - 1] == nums[right] {
171-
right += 1;
172-
}
173-
} else {
174-
left += 1;
175-
while left < right && nums[left - 1] == nums[left] {
176-
left += 1;
177-
}
178-
if left == right {
179-
right += 1;
180-
}
199+
if vis.contains(&(x + k)) {
200+
ans.insert(x);
181201
}
202+
vis.insert(x);
182203
}
183-
res
204+
ans.len() as i32
184205
}
185206
}
186207
```

‎solution/0500-0599/0532.K-diff Pairs in an Array/Solution.cpp

Lines changed: 10 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,16 @@
11
class Solution {
22
public:
33
int findPairs(vector<int>& nums, int k) {
4-
unordered_set<int> vis;
5-
unordered_set<int> ans;
6-
for (int& v : nums) {
7-
if (vis.count(v - k)) ans.insert(v - k);
8-
if (vis.count(v + k)) ans.insert(v);
9-
vis.insert(v);
4+
unordered_set<int> ans, vis;
5+
for (int x : nums) {
6+
if (vis.count(x - k)) {
7+
ans.insert(x - k);
8+
}
9+
if (vis.count(x + k)) {
10+
ans.insert(x);
11+
}
12+
vis.insert(x);
1013
}
1114
return ans.size();
1215
}
13-
};
16+
};

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