128. Longest Consecutive Sequence(C++)
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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.
你看到[100, 4, 200, 1, 3, 2]这个数组,首先你会看99或者101在不在这个数组里,发现数组没这两个数,那么100组成的连续序列长度仅为1。接着会看5或者3在不在数组里,会发现3存在,5不存在;紧接着会看2在不在….直到发现0不在。从而得到4组成的最长序列为4。哈希表查询时间复杂度为O(1)。每个元素只被查找一次,总时间复杂度为O(n)。
class Solution {public: int longestConsecutive(const vector<int>& nums) { unordered_map<int, bool> used; for (auto i : nums) used[i] = false; int longest = 0; for (auto i : nums) { if (used[i]) continue; int length = 1; used[i] = true; for (int j = i + 1; used.find(j) != used.end(); ++j) { used[j] = true; ++length; } for (int j = i - 1; used.find(j) != used.end(); --j) { used[j] = true; ++length; } longest = max(longest, length); } return longest; }};
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