128. 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://discuss.leetcode.com/topic/6148/my-really-simple-java-o-n-solution-accepted/2

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:

1. 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.
2. 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.

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);        }        else {            // duplicates            continue;        }    }    return res;}