Longest Consecutive Sequence

最新推荐文章于 2022-05-12 10:20:38 发布

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本文介绍了一种使用哈希映射解决最长连续序列问题的方法。通过维护一个哈希映射来跟踪序列长度，并在插入新元素时更新边界点的值，实现了高效查找最长连续序列。

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问题解法1：

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 0means 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). Please comment if you see something wrong. Thanks.

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;
}

用python 实现的相同代码：

class Solution(object):
    def longestConsecutive(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dict ={}
        res = 0
        for num in nums:
            if dict.get(num) == None:
                left = dict.get(num - 1)
                right = dict.get(num + 1)
                if left == None:
                    left = 0
                if right == None:
                    right = 0
                sum = left + right + 1
                dict[num] = sum;
                dict[num - left] = sum
                dict[num + right] = sum
                res = max(res, sum)
        return res;