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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