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.

一刷
題解：
用map<Integer, Integer>來存儲數字和對應的consecutive的長度。
比如{1,2,3,4,5}，就將長度信息5保存在1，5中。每次只更新boundary和數字n

public class Solution {
    public int longestConsecutive(int[] num) {
    int res = 0;
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int n : num) {
        if (!map.containsKey(n)) {
            int left = (map.containsKey(n - 1)) ? map.get(n - 1) : 0;
            int right = (map.containsKey(n + 1)) ? map.get(n + 1) : 0;
            // sum: length of the sequence n is in
            int sum = left + right + 1;
            map.put(n, sum);
//very important, in case, duplicate will 算出左邊的sum和右邊的sum, 其實算了重復的兩次
            
            // keep track of the max length 
            res = Math.max(res, sum);
            
            // extend the length to the boundary(s)
            // of the sequence
            // will do nothing if n has no neighbors
            map.put(n - left, sum);
            map.put(n + right, sum);
        }
    }
    return res;
}
}