leetcode 128最长连续序列

 

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<int>& nums) {
        sort(nums.begin(),nums.end());
        int res=0;
        for(int i=0;i<nums.size();i++){
            int step=0,len=1;
            while(i+step!=nums.size()-1&&nums[i+step+1]-nums[i+step]<=1){
                if(nums[i+step]+1==nums[i+step+1]) len++;
                step++;
            }
            res=max(res,len);
            i+=step;
        }
        return res;
    }
};

 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<int>& nums) {
        int len=nums.size();
        if(len<=1) return len;
        unordered_map<int,int> 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<int>& nums) {
        int len=nums.size();
        if(len<=1) return len;
        unordered_map<int,int> 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<int>& nums) {
        int res=0;
        unordered_set<int> 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table

solution 1: hashtable (key,len)
case1: no neighboors
h[num]=1;
case2: one neighboor
l=h[num-1] or r=h[num+1]
h[num]=h[num-1]=l+1 or h[num]=h[num+1]=r+1
case3: two neighboors
l=h[num-1]
r=h[num+1]
h[num]=h[num-1]=h[num+1]=l+r+1



*****/
class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int res=0;
        unordered_map<int,int> 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<int>& nums) {
        int res=0;
        unordered_set<int> 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;
    }
};

 

有道词典
solution 1: has ...
详细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;}};
posted @ 2019-05-16 17:24  Joel_Wang  阅读(434)  评论(0编辑  收藏  举报