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feat: add solutions to lc/lcof2 problem:Longest Consecutive Sequence
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‎lcof2/剑指 Offer II 119. 最长连续序列/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法 1：排序**
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设 res 表示连续序列的最大长度，t 表示当前合法连续序列的长度，初始时 `res = t = 1`。
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先排序数组，然后从下标 1 开始遍历数组，判断 `nums[i]` 与前一个数 `nums[i - 1]` 的大小关系：
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- 若 `nums[i] == nums[i - 1]`，直接跳过；
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- 若 `nums[i] - nums[i - 1] == 1`，说明是连续序列，t 自增，利用 `res = max(res, t)` 更新最大长度；
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- 否则 t 重置为 1，继续往下遍历。
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此方法时间复杂度 `O(nlogn)`，空间复杂度 `O(1)`。
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**方法 2：哈希表**
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设 res 表示连续序列的最大长度，初始为 0。哈希表 s 存放数组出现的每个元素。
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遍历数组，以当前遍历到的元素 `nums[i]` 做为起点，循环判断 `nums[i] + 1`，`nums[i] + 2` ... 是否存在 s 中，并不断更新连续序列的最大长度。
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在这个过程中，如果 `nums[i]`, `nums[i] + 1`, `nums[i + 2]`, ... 是一个连续序列，遍历下个元素 `nums[i] + 1` 时，其实无需再重复循环。因此，只需要判断 `nums[i] - 1` 是否在 s 中，是则直接跳过。
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此方法时间复杂度 `O(n)`，空间复杂度 `O(n)`。
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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 longestConsecutive(self, nums: List[int]) -> int:
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n = len(nums)
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if n < 2:
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return n
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nums.sort()
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res = t = 1
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for i in range(1, n):
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if nums[i] == nums[i - 1]:
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continue
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if nums[i] - nums[i - 1] == 1:
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t += 1
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res = max(res, t)
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else:
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t = 1
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return res
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```
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```python
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class Solution:
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def longestConsecutive(self, nums: List[int]) -> int:
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s, res = set(nums), 0
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for num in nums:
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if num - 1 not in s:
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t = 1
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next = num + 1
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while next in s:
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t += 1
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next += 1
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res = max(res, t)
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return res
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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 longestConsecutive(int[] nums) {
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int n = nums.length;
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if (n < 1) {
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return n;
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}
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Arrays.sort(nums);
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int res = 1, t = 1;
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for (int i = 1; i < n; ++i) {
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if (nums[i] == nums[i - 1]) {
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continue;
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}
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if (nums[i] - nums[i - 1] == 1) {
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t += 1;
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res = Math.max(res, t);
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} else {
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t = 1;
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}
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}
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return res;
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}
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}
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```
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```java
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class Solution {
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public int longestConsecutive(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int num : nums) {
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s.add(num);
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}
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int res = 0;
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for (int num : nums) {
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if (!s.contains(num - 1)) {
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int t = 1, next = num + 1;
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while (s.contains(next++)) {
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++t;
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}
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res = Math.max(res, t);
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}
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}
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return res;
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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 longestConsecutive(vector<int> &nums) {
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int n = nums.size();
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if (n < 2)
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return n;
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sort(nums.begin(), nums.end());
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int res = 1, t = 1;
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for (int i = 1; i < n; ++i)
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{
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if (nums[i] == nums[i - 1])
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continue;
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if (nums[i] - nums[i - 1] == 1)
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{
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++t;
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res = max(res, t);
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}
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else
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{
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t = 1;
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}
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}
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return res;
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}
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};
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```
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```cpp
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class Solution {
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public:
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int longestConsecutive(vector<int> &nums) {
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unordered_set<int> s;
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for (int num : nums)
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s.insert(num);
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int res = 0;
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for (int num : nums)
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{
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if (!s.count(num - 1))
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{
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int t = 1, next = num + 1;
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while (s.count(next++))
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++t;
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res = max(res, t);
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}
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}
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return res;
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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 longestConsecutive(nums []int) int {
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n := len(nums)
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if n < 2 {
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return n
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}
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sort.Ints(nums)
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res, t := 1, 1
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for i := 1; i < n; i++ {
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if nums[i] == nums[i-1] {
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continue
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}
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if nums[i]-nums[i-1] == 1 {
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t++
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res = max(res, t)
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}
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}
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return res
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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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### **Go**
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```go
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func longestConsecutive(nums []int) int {
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s := make(map[int]bool)
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for _, num := range nums {
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s[num] = true
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}
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res := 0
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for _, num := range nums {
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if !s[num-1] {
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t, next := 1, num+1
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for s[next] {
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next++
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t++
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}
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res = max(res, t)
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}
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}
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return res
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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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### **...**

‎lcof2/剑指 Offer II 119. 最长连续序列/Solution.cpp

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class Solution {
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public:
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int longestConsecutive(vector<int> &nums) {
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unordered_set<int> s;
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for (int num : nums)
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s.insert(num);
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int res = 0;
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for (int num : nums)
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{
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if (!s.count(num - 1))
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{
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int t = 1, next = num + 1;
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while (s.count(next++))
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++t;
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res = max(res, t);
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}
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}
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return res;
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}
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};

‎lcof2/剑指 Offer II 119. 最长连续序列/Solution.go

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func longestConsecutive(nums []int) int {
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s := make(map[int]bool)
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for _, num := range nums {
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s[num] = true
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}
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res := 0
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for _, num := range nums {
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if !s[num-1] {
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t, next := 1, num+1
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for s[next] {
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next++
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t++
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}
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res = max(res, t)
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}
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}
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return res
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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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}

‎lcof2/剑指 Offer II 119. 最长连续序列/Solution.java

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class Solution {
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public int longestConsecutive(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int num : nums) {
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s.add(num);
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}
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int res = 0;
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for (int num : nums) {
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if (!s.contains(num - 1)) {
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int t = 1, next = num + 1;
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while (s.contains(next++)) {
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++t;
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}
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res = Math.max(res, t);
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}
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}
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return res;
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}
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}

‎lcof2/剑指 Offer II 119. 最长连续序列/Solution.py

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class Solution:
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def longestConsecutive(self, nums: List[int]) -> int:
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s, res = set(nums), 0
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for num in nums:
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if num - 1 not in s:
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t = 1
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next = num + 1
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while next in s:
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t += 1
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next += 1
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res = max(res, t)
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return res

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