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128. longest consecutive sequence

admin 11月 13, 2020 0

Question

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

Your algorithm should run in O(n) complexity.

Analysis

Use a HashMap to record the longest sequence it can reach map.put(nums[i], longest)

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.

Complexity

Time: O(n)
Space: O(1)

Solution

public int (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;
            
            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;
}