128. Longest Consecutive Sequence

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.

空间换时间，用了 map.
map 咋声明？判断 key = n 是否在该 map 中咋弄？

map<int, int> map;
if(map.find(n) == map.end()){} //若 key = n 不在map中

本题关键点(思路很清晰)：
主要分 2 steps:

扫描数组, n 表示当前扫到的元素．

step1:
	若 key = n 不在 map 中, 就找n的左(n-1)、右(n+1)邻居, 然后sum = left + right + 1,
	再把 map[n] = sum;

step2:
	接着更新　这个连续序列的 边界(boundarys, boundarys, 重要的话说３遍！) 为 sum 值，　如：
	e.g [1,2,3,4,5], A[1] = A[5] = 5, except A[2,..,4]!
	若无　左　或　右　边界，则不受影响 respectively, 如下核心代码：
	map[n - left] = sum;
	map[n + right] = sum;

key = n 在 map 中的话，跳过本次循环，因为重复了，不必理会了．

人家想法，咱家代码：
(O(n)) time, (O(n)) extra space.

int longestConsecutive(vector<int>& A) {
	int res = 0;
	map<int, int> map;

	for (int n : A) {
		// 若 key = n 不在map中
		if (map.find(n) == map.end()) {
			// step 1: left, right issues
			int left = (map.find(n - 1) != map.end()) ? map[n - 1] : 0;
			int right = (map.find(n + 1) != map.end()) ? map[n + 1] : 0;
			int sum = left + right + 1;
			map[n] = sum;

			res = max(res, sum);

			// step 2: boundary(s) issues

			// update the length of
			// this consecutive sequence boundary(s)
			// e.g [1,2,3,4,5], A[1] = A[5] = 5, except A[2,..,4]!
			// 若无　左　或　右　边界，则不受影响 respectively.
			map[n - left] = sum;
			map[n + right] = sum;
		} else {
			// duplicates
			continue;
		}
	}
	return res;
}

还有牛人的超短的代码：
https://leetcode.com/problems/longest-consecutive-sequence/discuss/