[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;
    }