方法一：使用快排：

//排序法，时间O(nlogn)，使用STL，只是验证一下思想，非正解；class Solution {public:    int longestConsecutive(vector
     
      & nums) {        sort(nums.begin(),nums.end());        int res=0;        for(int i=0;i
     

 方法二：使用并查集如题所说达到O(n)

 

方法三：使用哈希表O(n)

//哈希表结合染色，建立一个哈希表，然后遍历之后计数每个元素周围所有相邻元素并染色，记录个数；O(n)复杂度class Solution {public:    int longestConsecutive(vector
     
      & nums) {        int len=nums.size();        if(len<=1) return len;        unordered_map
      
        m;        int res=0;        for(int n:nums)            m[n]=1;        for(int n:nums){            int i=n,j=n;            int cnt=1;            if(m[n]==0)                continue;            else                m[n]=0;            while(m[i+1]==1){                i++;                m[i]=0;            }                            while(m[j-1]==1){                j--;                m[j]=0;            }                            cnt=i-j+1;            res=cnt>res?cnt:res;        }        return res;    }};
      
     

别人家的哈希表：

/****通过哈希表记录边界信息neither i+1 nor i-1 has been seen: m[i]=1;both i+1 and i-1 have been seen: extend m[i+m[i+1]] and m[i-m[i-1]] to each other;only i+1 has been seen: extend m[i+m[i+1]] and m[i] to each other;only i-1 has been seen: extend m[i-m[i-1]] and m[i] to each other.*****/class Solution {public:    int longestConsecutive(vector
     
      & nums) {        int len=nums.size();        if(len<=1) return len;        unordered_map
      
        m;        int res=0;        for(int i:nums){            if(m[i]) continue;            //后面表达式为将m[i]和这一段连续序列的边界全部赋值为他的长度            //边界只可能为m[i-m[i-1]]到m[i+m[i+1]]，m[i]到m[i+m[i+1]],m[i-m[i-1]]到m[i]这几种情况，因此更新三者的值为新连续序列的长度即可            //又因为没有的元素哈希值为0，所以m[i]左右元素的m[i-1]+m[i+1]+1为新序列的长度；            res=max(res,m[i]=m[i+m[i+1]]=m[i-m[i-1]]=m[i-1]+m[i+1]+1);        }        return res;    }};
      
     

 别人家的使用hashset 和 hashtable：

hashset：

/****通过哈希set将nums转化为哈希set,然后对哈希set进行遍历，寻找连续片段的左边界，然后num+1进行遍历。可以证明每个元素将被访问2遍，for中一遍，while一遍，所以time O(n),space O(n);由于int会产生越界，可以使用long,也可以进行边界检测，INT_MAX break;*****/class Solution {public:    int longestConsecutive(vector
     
      & nums) {        int res=0;        unordered_set
      
        h(nums.begin(),nums.end());        for(int num:nums){            int l=0;            if(!h.count(num-1)){                while(h.count(num)!=0){                    ++l;                    if(num==INT_MAX) break;                    ++num;                }                res=res>l?res:l;            }        }        return res;    }};
      
     

hashtable：

/****通过哈希tablesolution 1: hashtable (key,len)case1: no neighboorsh[num]=1;case2: one neighboorl=h[num-1] or r=h[num+1]h[num]=h[num-1]=l+1 or h[num]=h[num+1]=r+1case3: two neighboorsl=h[num-1]r=h[num+1]h[num]=h[num-1]=h[num+1]=l+r+1*****/class Solution {public:    int longestConsecutive(vector
     
      & nums) {        int res=0;        unordered_map
      
        h;        for(int num:nums){            if(h[num]!=0) continue;            int l=h[num-1];            int r=h[num+1];                        int t=l+r+1;                        h[num]=h[num+r]=h[num-l]=t;            res=res>t?res:t;        }        return res;    }};
      
     

 哈希set

/****通过哈希set将nums转化为哈希set,然后对哈希set进行遍历，寻找连续片段的左边界，然后num+1进行遍历。可以证明每个元素将被访问2遍，for中一遍，while一遍，所以time O(n),space O(n);由于int会产生越界，可以使用long,也可以进行边界检测，INT_MAX break;*****/class Solution {public:    int longestConsecutive(vector
     
      & nums) {        int res=0;        unordered_set
      
        h(nums.begin(),nums.end());        for(int num: nums){            int l=0;            if(h.count(num-1)==0){                while(h.count(num)>0){                    l++;                    if(num==INT_MAX) break;                    num++;                }            }            res=res>l?res:l;        }        return res;    }};
      
     

 

solution 1: has ...

X

　　解决方案1:哈希表(关键,len) case1:没有neighboorsh (num) = 1,例2:一个neighboorl = h [num-1]或r = h (num + 1) h (num) = h [num-1] = l + 1或h (num) = h (num + 1) = r + 1 case3:两个neighboorsl = h [num-1] r = h (num + 1) h (num) = h [num-1] = h (num + 1) = l + r + 1 　　 　　 　　* * * * * /类解决方案{公众:int longestConsecutive(向量< int > & num) {int res = 0; unordered_map < int, int > h;为(int num: num){如果(h (num) ! = 0)继续;int l = h [num-1]; int r = h (num + 1); int t = l + r + 1; h (num) = h (num + r) = h [num-l] = t; res = res > t ? res: t;}返回res;}};