[leetcode]128. Longest Consecutive Sequence最长连续序列

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.

题目

给定一个数组，计算将其排序以后能形成的最长连续序列。

思路

如果允许O(nlogn)的复杂度，那么可以先排序。 以下是naive 的先排序的代码实现

 class Solution {
     public int longestConsecutive(int[] nums){
         Arrays.sort(nums);
         if(nums == null || nums.length == 0) return 0;
         int result = 1;
         int length = 1;

         for (int i = 1; i < nums.length; i++ ) {
             if(nums[i] == nums[i-1] + 1 ){
                 length ++;
             }else if (nums[i] == nums[i-1]){
                 continue;
             }else{
                 length = 1;

             }
             result = Math.max(result, length);
         }

      return  result;
     }
 }

再思考：

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)，想到用哈希set。

代码

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

         return max;
     }