【Leetcode】128. Longest Consecutive Sequence

题目：

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.

题解：

Solution 1

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {int longest_len = 0;
        unordered_map<int, bool> map;
        for(auto n : nums)
            map[n] = false;
        for(auto i : nums){
            if(map[i])
                continue;
            int len = 1;
            for(int j = i + 1; map.find(j) != map.end(); ++j){
                map[j] = true;
                ++len;
            }
            for(int j = i - 1; map.find(j) != map.end(); --j){
                map[j] = true;
                ++len;
            }
            longest_len = max(longest_len, len);
        }
        return longest_len;
    }
};

Solution 2

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int longest_len = 0;
        unordered_set<int> set(nums.begin(), nums.end());
        for(auto i : nums){
            if(set.find(i) == set.end())
                continue;
            set.erase(i);
            int prev = i - 1, next = i + 1;
            while(set.find(prev) != set.end()) set.erase(prev--);
            while(set.find(next) != set.end()) set.erase(next++);
            longest_len = max(longest_len, next - prev - 1);
        }
        return longest_len;
    }
};

Solution 3

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int longest_len = 0;
        unordered_map<int, int> map;
        for(auto n : nums){
            if(map.find(n) != map.end())
                continue;
            int prevsum = map.find(n - 1) != map.end() ? map[n - 1] : 0;
            int nextsum = map.find(n + 1) != map.end() ? map[n + 1] : 0;
            int sum = prevsum + nextsum + 1;
            map[n] = sum;
            map[n - prevsum] = sum;
            map[n + nextsum] = sum;
            longest_len = max(sum, longest_len);
        }
        return longest_len;
    }
};