Longest Consecutive Sequence

Longest Consecutive Sequence

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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.

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 updateint 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;}