Longest Consecutive Sequence

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.

思路

先对数组排序，然后用DP

public class Solution {
    public int longestConsecutive(int[] num) {
        Arrays.sort(num);
        int curLength = 1;
        int maxLength = 1;
        
        for(int i = 1; i < num.length; i++){
            if(num[i] == num[i - 1])
                continue;
            else if(num[i] - num[i - 1] == 1){
                curLength ++;
                maxLength = maxLength > curLength ? maxLength : curLength;
            }
            else
                curLength = 1;
        }//for
        
        return maxLength;
    }
}

 上面思路时间复杂度为O(N*logN)

题目要求用O(n)这里可以借助hashtable或者hashmap，先将数组全部放到hash表中，在遍历数组，找到前面和后面的数，这里利用了hash通过get方法时间复杂度为o(1)

import java.util.Hashtable;

public class Solution {
    public int longestConsecutive(int[] num) {
        Hashtable<Integer, Integer> map = new Hashtable<Integer, Integer>();
        for(int i = 0; i < num.length; i++){
            map.put(num[i], 1);
        }//放到hash表中
        
        int result = 0;
        for(int i = 0; i < num.length; i++){
            int sum = 1;
            if(map.get(num[i]) != null){
                int left = num[i] - 1;
                map.remove(num[i]);
                while(map.get(left) != null){
                    sum ++;
                    map.remove(left);
                    left--;
                }//while
                int right = num[i] + 1;
                while(map.get(right) != null){
                    sum++;
                    map.remove(right);
                    right++;
                }//while
                
                result = result > sum ? result : sum;
            }//if
        }//for
        
        return result;
    }
}