leetcode 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.

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这个题目，由于时间的限制，不能用暴力来做。可以用map+标记完成。

1.首先把所有的数字都放入到hash map当中，这样就能够迅速访问所有的数字。

2.接着开始扫描数组，每次获取数组的一个数字，如果数字没有被访问过，那么分别向左和向右延伸到头，并计算连续数字的长度。在延伸

过程中访问到的数字，都标记为已访问，下次再数组当中访问到该数字，就不需要延伸了。

3.复杂度,map是O(log(n)),所有访问一遍是n*log(n),然后再延续一遍是2*n*log(n).所以总的也是O(n*log(n))。至于用hashMap，使得hash的过程中，复杂度是O(1)，那么就符合题目要求了。

/**
 * Because we must find the consecutive numbers, randomly find a number, and extend it by one,
 * to see whether it is consecutive on the larger end and little end. and add the number of 
 * the larger extends and little extends. and then compare it to the max consecutive number.
 * 
 */
class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        set<int> nums;
		set<int>::iterator ite;
		int res,res1,val;
		// add all the number in set;
		for(int i=0;i<num.size();i++)
		{
			nums.insert(num[i]);
		}
		res = 0;
		// scan the array, to find some consecutive sequence
		for(int i=0;i<num.size();i++)
		{
			ite = nums.find(num[i]);
			if(ite!=nums.end())
			{
				res1 = 0;
				val = num[i];
				//search towards the small numbers
				while((ite=nums.find(val))!=nums.end())
				{
					res1++;
					nums.erase(val);
					val--;
				}
				val = num[i]+1;
				//search towards the large numbers
				while((ite=nums.find(val))!=nums.end())
				{
					res1++;
					nums.erase(val);
					val++;
				}
				res = max(res,res1);
			}
		}
		return res;
    }
};