Leetcode_128_Longest Consecutive Sequence

本文是在学习中的总结，欢迎转载但请注明出处：http://blog.csdn.net/pistolove/article/details/43854597

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.

思路：

（1）题意为给定一个未排序的数组，求数组中最长连续元素的个数。

（2）由于数组是未排序的，而需要求得数组中连续序列元素的个数。首先想到的应该是对数组进行排序，本文用的是java类库中自带的排序算法：Arrays.sort()。然后遍历数组，由于数组中可能出现多个连续序列，所以设置当前最长序列个数max和正在遍历的连续序列的个数count。首先，从数组下标为0开始往后遍历，判断当前元素和后续元素是否相差1，如果相差1，则count++，当遍历到倒数第二个元素时，返回count和max的较大值；如果相等，判断max和count大小，将较大值赋给max，继续遍历；其余情况为值相差大于1，这时将max和count较大值赋给max，并将count置为1（因为连续序列在这里断开了，需要重新记录）；遍历完整个数组，所得max即为最长连续序列个数。

（3）本文算法效率不是很高，有待后续优化。希望本文对你有所帮助。

算法代码实现如下：

/*** @author liqq*/
public class Longest_Consecutive_Sequence {public int longestConsecutive(int[] num) {if (num == null)  return -1;if (num.length == 1)  return 1;Arrays.sort(num);int count = 1;int max = 1;for (int i = 0; i < num.length - 1; i++) {if (num[i] + 1 == num[i + 1]) {count = count + 1;if (i == num.length - 2) {return count > max ? count : max;}} else if (num[i] == num[i + 1]) {max = count > max ? count : max;continue;} else {max = max > count ? max : count;count = 1;}}return max;}
}

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