LeetCode之Longest Consecutive Sequence

Descrition：
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.

思路：
主要通過unordered_map()函數來記錄已經訪問的數，話說unordered_map()函數是真方便啊。

#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<vector>
#include<string>
using namespace std;

class Solution{
public:
    int MaxLength(const vector<int>& matrix) {
        unordered_map<int, bool> used;
        if (matrix.empty()) return 0;
        for (auto i : matrix) used[i] = false;

        int max_length = 0;
        for (auto i:matrix){
            if (used[i]) continue;
            int length = 1;
            used[i] = true;

            //unordered_map<int,bool>
            for (int x = i + 1; used.find(x) != used.end(); x++) {
                used[x] = true;
                length++;
            }
            //if (j == 0) continue;
            for (int y = i - 1; used.find(y) != used.end(); y--) {
                used[y] = true;
                length++;

            }
            max_length = max(max_length,length);
        }
        return max_length;
    }


}; 
int main(int argc, char** argv) {

    vector<int> matrix;
    int temp=0;
    //string str;
    //cin >> str;
    cout << "Please input matrix:" << endl;
    for (int i = 0; i < 10; i++) {
        cin >> temp;
        matrix.push_back(temp);
    }
    Solution test;
    cout<<"Result:"<<test.MaxLength(matrix)<<endl;

    //unordered_map<string, double> mymap = { {"wang",1},{"chao",2},{"long",3} };

    //unordered_map<string, double>::iterator itr = mymap.find(str);
    //
    //if (itr != mymap.end())
    //  cout << itr->first << " is " << itr->second << endl;
    //else
    //{
    //  cout << "no found!" << endl;
    //}
    system("pause");
    return 0;
}

結果如下：