[leetcode] 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.

https://oj.leetcode.com/problems/longest-consecutive-sequence/

 

思路：用hashset存储所有元素，对在hashset中某一元素，在hashset中删除，然后向前向后依次寻找还在hashset中的元素并删除，期间更新最大值。

 

 

public class Solution {

    public int longestConsecutive(int[] num) {
        if (num == null || num.length == 0)
            return 0;
        if (num.length == 1)
            return 1;
        HashSet<Integer> set = new HashSet<Integer>();
        for (int each : num)
            set.add(each);

        int res = 0;

        for (int i = 0; i < num.length; i++) {
            int count = 0;
            if (set.contains(num[i])) {
                count++;
                set.remove(num[i]);
                int pre = num[i] - 1;
                while (set.contains(pre)) {
                    set.remove(pre);
                    count++;
                    pre--;
                }
                int post = num[i] + 1;
                while (set.contains(post)) {
                    set.remove(post);
                    count++;
                    post++;
                }

            }
            if (count > res)
                res = count;

        }
        return res;

    }

    public static void main(String[] args) {

        System.out.println(new Solution().longestConsecutive(new int[] { 1,2,4,5,6 }));
    }

}

 

第二遍记录：

　　注意遍历的时候，是遍历num不是set，因为num可能有重复元素，size可能会小很多。

public class Solution {

    public int longestConsecutive(int[] num) {
        if(num==null||num.length==0)
            return 0;
        HashSet<Integer> set = new HashSet<Integer>();
        for(int i=0;i<num.length;i++)
            set.add(num[i]);
        
        int res =1;
        for(int i=0;i<num.length;i++){
            int count = 0;
            if(set.contains(num[i])){
                count++;
                set.remove(num[i]);
                int pre = num[i]-1;
                while(set.contains(pre)){
                    count++;
                    set.remove(pre);
                    pre--;
                }
                int post =num[i]+1;
                while(set.contains(post)){
                    count++;
                    set.remove(post);
                    post++;
                }
                res=Math.max(res,count);
            }
            
        }
        
        return res;
    }

}

 

 

参考：

http://blog.csdn.net/fightforyourdream/article/details/15024861