算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Longest Consecutive Sequence


### 描述

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
Given `[100, 4, 200, 1, 3, 2]`,
The longest consecutive elements sequence is `[1, 2, 3, 4]`. Return its length: 4.

Your algorithm should run in `O(n)` complexity.


### 分析

如果允许$$O(n \log n)$$的复杂度,那么可以先排序,可是本题要求`O(n)`。

由于序列里的元素是无序的,又要求`O(n)`,首先要想到用哈希表。

用一个哈希表存储所有出现过的元素,对每个元素,以该元素为中心,往左右扩张,直到不连续为止,记录下最长的长度。


### 代码

{% if book.java %}
```java
// Longest Consecutive Sequence
// Time Complexity: O(n),Space Complexity: O(n)
public class Solution {
public int longestConsecutive(int[] nums) {
final HashSet mySet = new HashSet();
for (int i : nums) mySet.add(i);

int longest = 0;
for (int i : nums) {
int length = 1;
for (int j = i - 1; mySet.contains(j); --j) {
mySet.remove(j);
++length;
}
for (int j = i + 1; mySet.contains(j); ++j) {
mySet.remove(j);
++length;
}
longest = Math.max(longest, length);
}
return longest;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Longest Consecutive Sequence
// Time Complexity: O(n),Space Complexity: O(n)
class Solution {
public:
int longestConsecutive(const vector &nums) {
unordered_set my_set;
for (auto i : nums) my_set.insert(i);

int longest = 0;
for (auto i : nums) {
int length = 1;
for (int j = i - 1; my_set.find(j) != my_set.end(); --j) {
my_set.erase(j);
++length;
}
for (int j = i + 1; my_set.find(j) != my_set.end(); ++j) {
my_set.erase(j);
++length;
}
longest = max(longest, length);
}
return longest;
}
};
```
{% endif %}