[LeetCode] Longest Consecutive Sequence（DP）

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1：显然运行不是O（n）的时间复杂度，因此Time Limit Exceeded！

class Solution {
public:
int longestConsecutive(vector<int> &num) {
int len = num.size();
if(len<=1)
return len;
int max = 1;
map<int,int> m;//key is the number in num,value is the longest consecutive number from the  current key to  the bigger
for(int i=0;i<len;i++){

if(m.empty())
m[num[i]] = 1;
else if(m.count(num[i]+1)!= 0)
{
int number = num[i];
m[num[i]] = m[num[i]+1]+1;

while(m.count(number-1)!= 0)
{
m[number-1] = m[number]+1;
number--;
}
max = max>m[number] ? max :m[number];
}
else{
int number = num[i];
m[num[i]] = 1;
while(m.count(number-1)!= 0)
{
m[number-1] = m[number]+1;
number--;
}
max = max>m[number] ? max :m[number];
}

}//end for
return max;
}
};

方法2：其实和方法1一样的思想，只是用了map<int,vector<int>::iterator>来存储每个元素如果连续的话的上界或者下界，大大简化了
方法1中的2个while循环，这就是方法2改进的地方了。

class Solution {
public:
int longestConsecutive(vector<int> &num) {
map<int,int> vTable;//v(x) = the max length of consecutive sequence starting from x
map<int,vector<int>::iterator> aTable;
for (vector<int>::iterator i = num.begin(); i!=num.end(); i++) {
if(vTable.count(*i)) continue;      // Ignore same number
vTable[*i]=1;
aTable[*i]=i;                       // Initialization of new input
if(vTable.count(*i+1)) {            // If i+1 exists
vTable[*i] += vTable[*i+1];     // Update v(x)
aTable[*i] = aTable[*i+1];      // Update a(x)
}
if(vTable.count(*i-1)) {            // If i-1 exists, same idea
vTable[*aTable[*i-1]] += vTable[*i];
aTable[*aTable[*i]] = aTable[*i-1];
aTable[*aTable[*i-1]] = (vTable.count(*i+1)) ? aTable[*i] : i;
}else aTable[*aTable[*i]] = i;
}
int max=0; // Find max in vTable
map<int,int>::iterator iter ;
for (iter = vTable.begin();iter!= vTable.end();iter++)
if ((*iter).second>max)
max = (*iter).second;
return max;
}
};