zoukankan      html  css  js  c++  java
  • [LeetCode] Longest Consecutive Sequence

    This problem is not very intuitive at first glance. However, the final idea should be very self-explanatory. You visit each element in nums, and then find its left and right neighbors and extend the length accordingly. The time complexity is O(n) if we guard from visiting the same element again.

    You can implement it using an unordered_set or unordered_map.

    The code is as follows. The last one is taken from this page which includes a nice explanation in the follong answers.

     1 // O(n) solution using unordered_set
     2 int longestConsecutive(vector<int>& nums) {
     3     unordered_set<int> copy(nums.begin(), nums.end());
     4     unordered_set<int> filter;
     5     int len = 0;
     6     for (int i = 0; i < nums.size(); i++) {
     7         if (filter.find(nums[i]) != filter.end()) continue;
     8         int l = nums[i] - 1, r = nums[i] + 1;
     9         while (copy.find(l) != copy.end()) filter.insert(l--);
    10         while (copy.find(r) != copy.end()) filter.insert(r++);
    11         len = max(len, r - l - 1);
    12     }
    13     return len;
    14 }
    15 
    16 // O(n) solution using unordered_map
    17 int longestConsecutive(vector<int>& nums) {
    18     unordered_map<int, int> mp;
    19     int len = 0;
    20     for (int i = 0; i < nums.size(); i++) {
    21         if (mp[nums[i]]) continue;
    22         mp[nums[i]] = 1;
    23         if (mp.find(nums[i] - 1) != mp.end())
    24             mp[nums[i]] += mp[nums[i] - 1];
    25         if (mp.find(nums[i] + 1) != mp.end())
    26             mp[nums[i]] += mp[nums[i] + 1];
    27         mp[nums[i] - mp[nums[i] - 1]] = mp[nums[i]]; // left boundary
    28         mp[nums[i] + mp[nums[i] + 1]] = mp[nums[i]]; // right boundary
    29         len = max(len, mp[nums[i]]);
    30     }
    31     return len;
    32 }
    33 
    34 // O(n) super-concise solution (merging the above cases)
    35 int longestConsecutive(vector<int>& nums) {
    36     unordered_map<int, int> mp;
    37     int len = 0;
    38     for (int i = 0; i < nums.size(); i++) {
    39         if (mp[nums[i]]) continue;
    40         len = max(len, mp[nums[i]] = mp[nums[i] - mp[nums[i] - 1]] = mp[nums[i] + mp[nums[i] + 1]] = mp[nums[i] - 1] + mp[nums[i] + 1] + 1);
    41     }
    42     return len;
    43 }
  • 相关阅读:
    java.sql.SQLException: The server time zone value 'Öйú±ê׼ʱ¼ä'.. 问题解决方法
    Mysql存储引擎federated
    实习心语
    Linux版本CentOS、Ubuntu和Debian的异同
    Ubuntu忘记MySQL密码重设方法
    运行时异常和一般异常
    网络爬虫-正方教务系统登录
    大四心语
    缓存更新的套路
    (String)、toString、String.valueOf的区别
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4572028.html
Copyright © 2011-2022 走看看