Longest Consecutive Sequence

Longest Consecutive Sequence

  Total Accepted: 6466  Total Submissions: 24260 My Submissions

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.

3到4星级难度。

考点：活用hash表

利用?:操作可以简化让程序更加简洁。

注意：

1 数组中的数字会有重复，不要重复处理数据。

2 利用hash数据结构存储新增的数字，更新连续数字两边的边界值。

3 统计hash表中的最大值。

int longestConsecutive(vector<int> &num) 
	{
		unordered_map<int, int> mii;
		int len = 0;

		for (int i = 0; i < num.size(); i++)
		{
			if (!mii.count(num[i]))
			{
				int a = mii.count(num[i]-1)? mii[num[i]-1]:0;
				int b = mii.count(num[i]+1)? mii[num[i]+1]:0;
				mii[num[i]] = a+b+1;
				if(b) mii[num[i]+b] = a+b+1;
				if(a) mii[num[i]-a] = a+b+1;
				len = max(len, a+b+1);
			}
		}
		return len;
	}

//2014-2-18 update
	int longestConsecutive(vector<int> &num) 
	{
		unordered_map<int, int> ump_ii;
		int ans = 1;
		for (int i = 0; i < num.size(); i++)
		{
			if (ump_ii.count(num[i])) continue;
			ump_ii[num[i]] = 1;
			int pos1 = 0;
			if (ump_ii.count(num[i]+1))
			{
				pos1 = ump_ii[num[i]+1];
				ump_ii[num[i]] = pos1+1;
				ump_ii[num[i]+pos1] = ump_ii[num[i]];
			}
			if (ump_ii.count(num[i]-1))
			{
				int pos2 = ump_ii[num[i]-1];
				ump_ii[num[i]] += pos2;
				ump_ii[num[i]+pos1] = ump_ii[num[i]-pos2] = ump_ii[num[i]];
			}
			ans = max(ans, ump_ii[num[i]]);
		}
		return ans;
	}