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# 128.Longest-Consecutive-Sequence

## 题目描述

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
Your algorithm should run in O(n) complexity.
Example:
Input: [100, 4, 200, 1, 3, 2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

## 代码

### Approach #1 One Pass

public class Solution {
public int longestConsecutive(int[] nums) {
if (nums.length == 0) return 0;
Arrays.sort(nums);
int longestStreak = 1;
int currentStreak = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] != nums[i - 1]) {
if (nums[i] == nums[i - 1] + 1){
currentStreak += 1;
} else {
longestStreak = Math.max(longestStreak, currentStreak);
currentStreak = 1;
}
}
}
return Math.max(longestStreak, currentStreak);
}
}

### Approah 2: HashSet and Intelligent Sequence Building

public class Solution {
public int longestCOnsecutive(int[] nums) {
Set<Integer> numSet = new HashSet<Integer>();
for (int num : nums) {
}
int longestStreak = 0;
for (int num : numSet) {
if (!numSet.contains(num - 1)) {
int currentNum = num;
int currentStreak = 1;
while (numSet.contains(currentNum + 1)) {
currentNum += 1;
currentStreak += 1;
}
longestStreak = Math.max(longestStreak, currentStreak);
}
}
return longestSteak;
}
}