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feat: add solutions to lc problems: No.0961,0963 (doocs#1497)
* No.0961.N-Repeated Element in Size 2N Array * No.0963.Minimum Area Rectangle II
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‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/README.md

+50-22
Original file line numberDiff line numberDiff line change
@@ -58,9 +58,13 @@
5858

5959
<!-- 这里可写通用的实现逻辑 -->
6060

61-
长度为 `2N`，共 `N+1` 个不同元素，其中一个元素出现 `N` 次，说明其它元素各不相同。
61+
**方法一：哈希表**
6262

63-
遍历数组，只要出现重复元素，它就是我们要找的重复 `N` 次的元素。
63+
由于数组 $nums$ 一共有 $2n$ 个元素，其中有 $n + 1$ 个不同的元素，且有一个元素重复了 $n$ 次，说明数组中的其余 $n$ 个元素都是不同的。
64+
65+
因此，我们只需要遍历数组 $nums$，用哈希表 $s$ 记录遍历过的元素。当遍历到某个元素 $x$ 时，如果 $x$ 在哈希表 $s$ 中已经存在，说明 $x$ 是重复的元素，直接返回 $x$ 即可。
66+
67+
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
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6569
<!-- tabs:start -->
6670

@@ -72,10 +76,10 @@
7276
class Solution:
7377
def repeatedNTimes(self, nums: List[int]) -> int:
7478
s = set()
75-
for num in nums:
76-
if num in s:
77-
return num
78-
s.add(num)
79+
for x in nums:
80+
if x in s:
81+
return x
82+
s.add(x)
7983
```
8084

8185
### **Java**
@@ -85,14 +89,12 @@ class Solution:
8589
```java
8690
class Solution {
8791
public int repeatedNTimes(int[] nums) {
88-
Set<Integer> s = new HashSet<>();
89-
for (int num : nums) {
90-
if (s.contains(num)) {
91-
return num;
92+
Set<Integer> s = new HashSet<>(nums.length / 2 + 1);
93+
for (int i = 0;; ++i) {
94+
if (!s.add(nums[i])) {
95+
return nums[i];
9296
}
93-
s.add(num);
9497
}
95-
return -1;
9698
}
9799
}
98100
```
@@ -104,17 +106,44 @@ class Solution {
104106
public:
105107
int repeatedNTimes(vector<int>& nums) {
106108
unordered_set<int> s;
107-
for (auto& num : nums) {
108-
if (s.find(num) != s.end()) {
109-
return num;
109+
for (int i = 0;; ++i) {
110+
if (s.count(nums[i])) {
111+
return nums[i];
110112
}
111-
s.insert(num);
113+
s.insert(nums[i]);
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}
113-
return -1;
114115
}
115116
};
116117
```
117118
119+
### **Go**
120+
121+
```go
122+
func repeatedNTimes(nums []int) int {
123+
s := map[int]bool{}
124+
for i := 0; ; i++ {
125+
if s[nums[i]] {
126+
return nums[i]
127+
}
128+
s[nums[i]] = true
129+
}
130+
}
131+
```
132+
133+
### **TypeScript**
134+
135+
```ts
136+
function repeatedNTimes(nums: number[]): number {
137+
const s: Set<number> = new Set();
138+
for (const x of nums) {
139+
if (s.has(x)) {
140+
return x;
141+
}
142+
s.add(x);
143+
}
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}
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```
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118147
### **JavaScript**
119148

120149
```js
@@ -124,13 +153,12 @@ public:
124153
*/
125154
var repeatedNTimes = function (nums) {
126155
const s = new Set();
127-
for (const num of nums) {
128-
if (s.has(num)) {
129-
return num;
156+
for (const x of nums) {
157+
if (s.has(x)) {
158+
return x;
130159
}
131-
s.add(num);
160+
s.add(x);
132161
}
133-
return -1;
134162
};
135163
```
136164

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/README_EN.md

+44-20
Original file line numberDiff line numberDiff line change
@@ -45,25 +45,23 @@
4545
class Solution:
4646
def repeatedNTimes(self, nums: List[int]) -> int:
4747
s = set()
48-
for num in nums:
49-
if num in s:
50-
return num
51-
s.add(num)
48+
for x in nums:
49+
if x in s:
50+
return x
51+
s.add(x)
5252
```
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5454
### **Java**
5555

5656
```java
5757
class Solution {
5858
public int repeatedNTimes(int[] nums) {
59-
Set<Integer> s = new HashSet<>();
60-
for (int num : nums) {
61-
if (s.contains(num)) {
62-
return num;
59+
Set<Integer> s = new HashSet<>(nums.length / 2 + 1);
60+
for (int i = 0;; ++i) {
61+
if (!s.add(nums[i])) {
62+
return nums[i];
6363
}
64-
s.add(num);
6564
}
66-
return -1;
6765
}
6866
}
6967
```
@@ -75,17 +73,44 @@ class Solution {
7573
public:
7674
int repeatedNTimes(vector<int>& nums) {
7775
unordered_set<int> s;
78-
for (auto& num : nums) {
79-
if (s.find(num) != s.end()) {
80-
return num;
76+
for (int i = 0;; ++i) {
77+
if (s.count(nums[i])) {
78+
return nums[i];
8179
}
82-
s.insert(num);
80+
s.insert(nums[i]);
8381
}
84-
return -1;
8582
}
8683
};
8784
```
8885
86+
### **Go**
87+
88+
```go
89+
func repeatedNTimes(nums []int) int {
90+
s := map[int]bool{}
91+
for i := 0; ; i++ {
92+
if s[nums[i]] {
93+
return nums[i]
94+
}
95+
s[nums[i]] = true
96+
}
97+
}
98+
```
99+
100+
### **TypeScript**
101+
102+
```ts
103+
function repeatedNTimes(nums: number[]): number {
104+
const s: Set<number> = new Set();
105+
for (const x of nums) {
106+
if (s.has(x)) {
107+
return x;
108+
}
109+
s.add(x);
110+
}
111+
}
112+
```
113+
89114
### **JavaScript**
90115

91116
```js
@@ -95,13 +120,12 @@ public:
95120
*/
96121
var repeatedNTimes = function (nums) {
97122
const s = new Set();
98-
for (const num of nums) {
99-
if (s.has(num)) {
100-
return num;
123+
for (const x of nums) {
124+
if (s.has(x)) {
125+
return x;
101126
}
102-
s.add(num);
127+
s.add(x);
103128
}
104-
return -1;
105129
};
106130
```
107131

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/Solution.cpp

+11-12
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,12 @@
1-
class Solution {
2-
public:
3-
int repeatedNTimes(vector<int>& nums) {
4-
unordered_set<int> s;
5-
for (auto& num : nums) {
6-
if (s.find(num) != s.end()) {
7-
return num;
8-
}
9-
s.insert(num);
10-
}
11-
return -1;
12-
}
1+
class Solution {
2+
public:
3+
int repeatedNTimes(vector<int>& nums) {
4+
unordered_set<int> s;
5+
for (int i = 0;; ++i) {
6+
if (s.count(nums[i])) {
7+
return nums[i];
8+
}
9+
s.insert(nums[i]);
10+
}
11+
}
1312
};

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/Solution.go

+9
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,9 @@
1+
func repeatedNTimes(nums []int) int {
2+
s := map[int]bool{}
3+
for i := 0; ; i++ {
4+
if s[nums[i]] {
5+
return nums[i]
6+
}
7+
s[nums[i]] = true
8+
}
9+
}

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/Solution.java

+9-11
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,10 @@
1-
class Solution {
2-
public int repeatedNTimes(int[] nums) {
3-
Set<Integer> s = new HashSet<>();
4-
for (int num : nums) {
5-
if (s.contains(num)) {
6-
return num;
7-
}
8-
s.add(num);
9-
}
10-
return -1;
11-
}
1+
class Solution {
2+
public int repeatedNTimes(int[] nums) {
3+
Set<Integer> s = new HashSet<>(nums.length / 2 + 1);
4+
for (int i = 0;; ++i) {
5+
if (!s.add(nums[i])) {
6+
return nums[i];
7+
}
8+
}
9+
}
1210
}

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/Solution.js

+4-5
Original file line numberDiff line numberDiff line change
@@ -4,11 +4,10 @@
44
*/
55
var repeatedNTimes = function (nums) {
66
const s = new Set();
7-
for (const num of nums) {
8-
if (s.has(num)) {
9-
return num;
7+
for (const x of nums) {
8+
if (s.has(x)) {
9+
return x;
1010
}
11-
s.add(num);
11+
s.add(x);
1212
}
13-
return -1;
1413
};

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/Solution.py

+7-7
Original file line numberDiff line numberDiff line change
@@ -1,7 +1,7 @@
1-
class Solution:
2-
def repeatedNTimes(self, nums: List[int]) -> int:
3-
s = set()
4-
for num in nums:
5-
if num in s:
6-
return num
7-
s.add(num)
1+
class Solution:
2+
def repeatedNTimes(self, nums: List[int]) -> int:
3+
s = set()
4+
for x in nums:
5+
if x in s:
6+
return x
7+
s.add(x)

‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/Solution.ts

+9
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,9 @@
1+
function repeatedNTimes(nums: number[]): number {
2+
const s: Set<number> = new Set();
3+
for (const x of nums) {
4+
if (s.has(x)) {
5+
return x;
6+
}
7+
s.add(x);
8+
}
9+
}

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