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feat: update solutions to lc problem: No.0961. N-Repeated Element in Size 2N Array
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‎solution/0900-0999/0961.N-Repeated Element in Size 2N Array/README.md

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Original file line numberDiff line numberDiff line change
@@ -51,22 +51,80 @@
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<!-- 这里可写通用的实现逻辑 -->
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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。
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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。
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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 repeatedNTimes(self, nums: List[int]) -> int:
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s = set()
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for num in nums:
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if num in s:
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return num
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s.add(num)
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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 repeatedNTimes(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int num : nums) {
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if (s.contains(num)) {
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return num;
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}
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s.add(num);
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}
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return -1;
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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 repeatedNTimes(vector<int>& nums) {
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unordered_set<int> s;
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for (auto &num : nums) {
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if (s.find(num) != s.end()) {
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return num;
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}
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s.insert(num);
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}
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return -1;
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}
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};
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var repeatedNTimes = function(nums) {
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const s = new Set();
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for (const num of nums) {
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if (s.has(num)) {
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return num;
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}
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s.add(num);
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}
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return -1;
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};
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```
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### **...**

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

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@@ -58,13 +58,67 @@
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### **Python3**
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```python
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class Solution:
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def repeatedNTimes(self, nums: List[int]) -> int:
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s = set()
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for num in nums:
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if num in s:
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return num
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s.add(num)
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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 repeatedNTimes(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int num : nums) {
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if (s.contains(num)) {
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return num;
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}
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s.add(num);
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}
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return -1;
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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 repeatedNTimes(vector<int>& nums) {
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unordered_set<int> s;
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for (auto &num : nums) {
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if (s.find(num) != s.end()) {
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return num;
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}
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s.insert(num);
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}
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return -1;
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}
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};
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var repeatedNTimes = function(nums) {
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const s = new Set();
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for (const num of nums) {
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if (s.has(num)) {
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return num;
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}
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s.add(num);
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}
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return -1;
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};
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```
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### **...**

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

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Original file line numberDiff line numberDiff line change
@@ -1,12 +1,13 @@
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class Solution {
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public:
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int repeatedNTimes(vector<int>& A) {
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unordered_set<int> us(A[0]) ;
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for(int i = (A.size() >> 1) - 1; i < A.size(); ++i)
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if(us.find(A[i]) != us.end())
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return A[i] ;
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else
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us.insert(A[i]) ;
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return A[0] ;
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int repeatedNTimes(vector<int>& nums) {
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unordered_set<int> s;
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for (auto &num : nums) {
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if (s.find(num) != s.end()) {
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return num;
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}
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s.insert(num);
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}
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return -1;
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}
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};

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

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Original file line numberDiff line numberDiff line change
@@ -1,12 +1,12 @@
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class Solution {
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public int repeatedNTimes(int[] A) {
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Set<Integer> set = new HashSet<>();
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for (int e : A) {
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if (set.contains(e)) {
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return e;
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public int repeatedNTimes(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int num : nums) {
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if (s.contains(num)) {
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return num;
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}
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set.add(e);
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s.add(num);
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}
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return 0;
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return -1;
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}
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}

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

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Original file line numberDiff line numberDiff line change
@@ -1,10 +1,14 @@
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const repeatedNTimes = function (A) {
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let ss = new Set();
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for (let i = 0; i < A.length; i++) {
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if (ss.has(A[i])) {
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return A[i];
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}
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ss.add(A[i]);
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var repeatedNTimes = function(nums) {
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const s = new Set();
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for (const num of nums) {
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if (s.has(num)) {
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return num;
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}
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s.add(num);
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}
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return null;
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};
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return -1;
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};

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

+6-16
Original file line numberDiff line numberDiff line change
@@ -1,17 +1,7 @@
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class Solution:
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def repeatedNTimes(self, A):
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"""
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:type A: List[int]
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:rtype: int
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"""
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if A[0] == A[1] or A[0] == A[2] or A[0] == A[3]:
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return A[0]
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elif A[1] == A[2] or A[1] == A[3]:
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return A[1]
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elif A[2] == A[3]:
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return A[2]
13-
i = 4
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while i < len(A):
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if A[i] == A[i + 1]:
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return A[i]
17-
i += 2
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def repeatedNTimes(self, nums: List[int]) -> int:
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s = set()
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for num in nums:
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if num in s:
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return num
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s.add(num)

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