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.

Ideas:

We will use HashMap. The key thing is to keep track of the sequence length and store that in the boundary points of the sequence. For example, as a result, for sequence {1, 2, 3, 4, 5}, map.get(1) and map.get(5) should both return 5.

Whenever a new element n is inserted into the map, do two things:

• See if n - 1 and n + 1 exist in the map, and if so, it means there is an existing sequence next to n. Variables left and right will be the length of those two sequences, while 0 means there is no sequence and n will be the boundary point later. Store (left + right + 1) as the associated value to key n into the map.
• Use left and right to locate the other end of the sequences to the left and right of n respectively, and replace the value with the new length. Everything inside the for loop is O(1) so the total time is O(n).

这个其实算是一种使用HashMap的Union Find方法，如果n的左右邻居也在HashMap中的话，就把他们unite为n的sum。

Code:

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);

            // 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;
}

另外一种解法DFS

public int longestConsecutive(int[] num) {
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int i = 0; i < num.length; i++) {
        map.put(num[i], i);
    }

    boolean[] visited = new boolean[num.length];

    int max = 1;
    int count = 1;
    for (int i = 0; i < num.length; i++) {
        if (visited[i]) continue;

        visited[i] = true;
        count = 1;
        int index = num[i]-1;
        while(map.containsKey(index)) {
            visited[map.get(index)] = true;
            index--;
            count++;
        }

        index = num[i]+1;
        while(map.containsKey(index)) {
            visited[map.get(index)] = true;
            index++;
            count++;
        }

        if (count > max) max = count;
    }
    return max;        
}