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.
思路：本题要求复杂度为O(n)，并且注意如果数组中有相同的数，那么他们不算连续的数。这样，我们使用一个map来存储原数组，首先就消除了重复的数， 将数组中的值同时作为map的first以及second，这样遍历map是就是从小到大的顺序，不需要排序，只需要第一次遍历数组，为O(n)，第二次遍历map也为O(n)就可以了。
class Solution { public: int longestConsecutive(vector<int>& nums) { if (nums.size() == 0) { return 0; } map<int, int>hash; for (int i = 0; i < nums.size(); i++) { hash[nums[i]] = nums[i]; } map<int, int>::iterator it = hash.begin(); int count = 1; int max = 1; while ((++it) != hash.end()) { it--; if (it -> second == (++it) -> second - 1) { count++; } else { count = 1; } if (count > max) { max = count; } } return max; } };
