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[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; }};
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