[LeetCode] Longest Consecutive Sequence

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.

方法一

　　先排序，然后找就好了，但是要求是O(n),排序明显是个O(n*logn)的算法.

方法二

只是找连续的嘛,我们可以把所有的数字都存入hash表,然后随意从某个数字开始找他的前面和后面那个是否存在.

然后得到一个最大的长度.当然找过的就可以删掉了...你想,一个连续的序列,你从中间任意位置开始往两边找不都一样么.

所以只要找过就可以删掉了.

class Solution {
public:
    set<int> flag;
    int findBound(int n , bool asc){
        int ans = 0;
        set<int>::iterator iter;
        while((iter = flag.find(n)) != flag.end()){
            flag.erase(iter);
            ans ++;
            if(asc) 
                n-- ;
            else 
                n++;
        }
        return ans;
    }
    int longestConsecutive(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
             
        int ans = 0;
        flag.clear();
        for(int i = 0 ; i < num.size() ; i++)
           flag.insert(num[i]);
        for(int i = 0 ; i < num.size(); i++){
            ans = max(findBound(num[i],true) + findBound(num[i]+1,false) , ans); 
        }
        return ans;
    }
};

下面用hash代替了set，思想一样，看上去没有上面的简洁

class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        map<int, int> hmap;
        hmap.clear();
        int n = num.size();
        for(int i=0; i<n; i++){
            hmap.insert(pair<int, int>(num[i], i));
        }
        int ans=0, cnt=0;
        map<int, int>::iterator it;
        for(int i=0; i<num.size(); i++){
            int cur = num[i];
            it = hmap.find(num[i]);
            cnt = 1;
            if(it!=hmap.end()){
                map<int, int>::iterator iter;
                while(1){
                    iter = hmap.find(++cur);
                    if(iter==hmap.end())
                        break;
                    cnt++;    
                    hmap.erase(iter);
                }
                cur=num[i];
                while(1){
                    iter = hmap.find(--cur);
                    if(iter==hmap.end())
                        break;
                    cnt++;    
                    hmap.erase(iter);
                }
                if(ans<cnt)
                    ans = cnt;           
                
            }
            
        }
        return ans;
    }
};