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;
    }
};