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 1:最简单的方法是 先排序, 然后扫描一遍数组。。复杂度是 O(nlgn) 不符合要求
思路 2:用 hash表 记录 n 个元素。
然后扫描数组,如果当前元素在 Hash 表中存在,求其所在的最大连续区间(向左,向右延伸)在搜索的过程中 删除 Hash 表中对应的元素。
如果当前元素在 Hash 表中不存在,这个元素要么不存在,要么被 删除了, 跳过。
class Solution { public: int longestConsecutive(vector<int> &num) { unordered_set<int> visited(num.begin(), num.end()); int len(0), longest(0); for(vector<int>::iterator it = num.begin(); it != num.end(); ++it){ if(visited.find(*it) == visited.end()) continue; len = 0; //向后延伸 for(int i=*it; visited.find(i)!=visited.end(); ++i, ++len) visited.erase(i); //向前延伸 for(int i=*it-1; visited.find(i)!=visited.end(); --i, ++len) visited.erase(i); longest = max(longest, len); } return longest; } };
LeetCode----Longest Consecutive Sequence
原文:http://blog.csdn.net/shoulinjun/article/details/19249039