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.
这道题要求求最长连续序列,并给定了O(n)复杂度限制,我们的思路是,使用一个集合set存入所有的数字,然后遍历数组中的每个数字,如果其在集合中存在,那么将其移除,然后分别用两个变量pre和next算出其前一个数跟后一个数,然后在集合中循环查找,如果pre在集合中,那么将pre移除集合,然后pre再自减1,直至pre不在集合之中,对next采用同样的方法,那么next-pre-1就是当前数字的最长连续序列,更新res即可。代码如下:
C++ 解法一:
class Solution { public: int longestConsecutive(vector<int>& nums) { int res = 0; unordered_set<int> s(nums.begin(), nums.end()); for (int val : nums) { if (!s.count(val)) continue; s.erase(val); int pre = val - 1, next = val + 1; while (s.count(pre)) s.erase(pre--); while (s.count(next)) s.erase(next++); res = max(res, next - pre - 1); } return res; } };
Java 解法一:
public class Solution { public int longestConsecutive(int[] nums) { int res = 0; Set<Integer> s = new HashSet<Integer>(); for (int num : nums) s.add(num); for (int num : nums) { if (s.remove(num)) { int pre = num - 1, next = num + 1; while (s.remove(pre)) --pre; while (s.remove(next)) ++next; res = Math.max(res, next - pre - 1); } } return res; } }
我们也可以采用哈希表来做,刚开始哈希表为空,然后遍历所有数字,如果该数字不在哈希表中,那么我们分别看其左右两个数字是否在哈希表中,如果在,则返回其哈希表中映射值,若不在,则返回0,然后我们将left+right+1作为当前数字的映射,并更新res结果,然后更新d-left和d-right的映射值,参见代码如下:
C++ 解法二:
class Solution { public: int longestConsecutive(vector<int>& nums) { int res = 0; unordered_map<int, int> m; for (int d : nums) { if (!m.count(d)) { int left = m.count(d - 1) ? m[d - 1] : 0; int right = m.count(d + 1) ? m[d + 1] : 0; int sum = left + right + 1; m[d] = sum; res = max(res, sum); m[d - left] = sum; m[d + right] = sum; } } return res; } };
Java 解法二:
public class Solution { public int longestConsecutive(int[] nums) { int res = 0; Map<Integer, Integer> m = new HashMap<Integer, Integer>(); for (int num : nums) { if (m.containsKey(num)) continue; int left = m.containsKey(num - 1) ? m.get(num - 1) : 0; int right = m.containsKey(num + 1) ? m.get(num + 1) : 0; int sum = left + right + 1; m.put(num, sum); res = Math.max(res, sum); m.put(num - left, sum); m.put(num + right, sum); } return res; } }
本文转自博客园Grandyang的博客,原文链接:求最长连续序列[LeetCode] Longest Consecutive Sequence ,如需转载请自行联系原博主。