【LeetCode从零单刷】Longest Consecutive Sequence
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题目:
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.
解答;
O(n) 的时间复杂度看到,就猜到是不是用动态规划。然后想了很长时间找不到状态转移方程……
结果 Discuss 里面都是用容器来计算(O(n) 不一定是DP,也可以考虑容器)。因为这段序列可能是有重复元素的,所以使用 set 相关的容器可以避免容器中包含重复元素。
但是,set / map 是基于红黑树实现的,因此查找效率是 O(log n) 级别的;因此,可以使用基于 hashmap 的 unordered_set 容器,查找效率是常数级别。
设定一个容器,保存已经访问过的元素。每个元素遍历它 ++ 以及 -- 部分。
class Solution {public: int longestConsecutive(vector<int>& nums) { unordered_set<int> visited; unordered_set<int> all(nums.begin(), nums.end()); int max = 0; int count, tmp; int size = nums.size(); for(int i=0; i<size; i++) { if (visited.find(nums[i]) != visited.end()) continue; visited.insert(nums[i]); count = 1; tmp = nums[i]; while(all.find(tmp - 1) != all.end()) { count++; tmp--; visited.insert(tmp); } tmp = nums[i]; while(all.find(tmp + 1) != all.end()) { count++; tmp++; visited.insert(tmp); } if (count > max) max = count; } return max; }};
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