原题传送门

1. 题目描述

2. Solution 1

1、思路分析
使用HashMap来记录连续序列长度，并把结果保存到边界点。如，给定序列{1, 2, 3, 4, 5}, map.get(1) 与 map.get(5)均得到5。
当一个新元素n插入到map时，做下面两件事

1. 查询n-1 和 n+1是否存在在map中，若存在，则表示序列延伸至n。变量left、right分别表示上述两序列长度，其中0表示不存在序列。把n -> (left + right +1) 插入到map中。
2. 用left、right来定位新序列的边界点，并更新边界的长度。

2、代码实现

package Q0199.Q0128LongestConsecutiveSequence;

import java.util.HashMap;

/*
   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.
   Everything inside the for loop is O(1) so the total time is O(n).
 */
public class Solution {

    public int longestConsecutive(int[] nums) {
        int res = 0;
        HashMap<Integer, Integer> ranges = new HashMap<>();
        for (int n : nums) {
            if (ranges.containsKey(n)) continue;
            // 1. Find left and right num
            int left = ranges.getOrDefault(n - 1, 0);
            int right = ranges.getOrDefault(n + 1, 0);
            int sum = left + right + 1;
            ranges.put(n, sum);  // Keep each number in Map to de-duplicate
            res = Math.max(res, sum);

            // 2. Union by only updating boundary
            // Leave middle k-v dirty to avoid cascading update
            if (left > 0) ranges.put(n - left, sum);
            if (right > 0) ranges.put(n + right, sum);
        }
        return res;
    }
}

/*
漂亮的cpp解法
use a hash map to store boundary information of consecutive sequence for each element; there are 4 cases when a new
element i reached:
1) neither i+1 nor i-1 has been seen: m[i]=1;
2) both i+1 and i-1 have been seen: extend m[i+m[i+1]] and m[i-m[i-1]] to each other;
3) only i+1 has been seen: extend m[i+m[i+1]] and m[i] to each other;
4) only i-1 has been seen: extend m[i-m[i-1]] and m[i] to each other.

int longestConsecutive(vector<int> &num) {
    unordered_map<int, int> m;
    int r = 0;
    for (int i : num) {
        if (m[i]) continue;
        r = max(r, m[i] = m[i + m[i + 1]] = m[i - m[i - 1]] = m[i + 1] + m[i - 1] + 1);
    }
    return r;
}
 */

3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(n)

标签：map,right,Sequence,int,sequence,ranges,Longest,Consecutive,left
来源： https://www.cnblogs.com/junstat/p/16287136.html