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C#LeetCode刷题之#674-最长连续递增序列( Longest Continuous Increasing Subsequence)

时间:2022-08-09 10:06:20

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C#LeetCode刷题之#674-最长连续递增序列( Longest Continuous Increasing Subsequence)

问题

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给定一个未经排序的整数数组,找到最长且连续的的递增序列。

输入:[1,3,5,4,7]

输出:3

解释:最长连续递增序列是 [1,3,5], 长度为3。尽管 [1,3,5,7] 也是升序的子序列, 但它不是连续的,因为5和7在原数组里被4隔开。

输入:[2,2,2,2,2]

输出:1

解释:最长连续递增序列是 [2], 长度为1。

注意:数组长度不会超过10000。

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

Input:[1,3,5,4,7]

Output:3

Explanation:The longest continuous increasing subsequence is [1,3,5], its length is 3.Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4.

Input:[2,2,2,2,2]

Output:1

Explanation:The longest continuous increasing subsequence is [2], its length is 1.

Note:Length of the array will not exceed 10,000.

示例

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public class Program {public static void Main(string[] args) {int[] nums = null;nums = new int[] { 1, 3, 5, 7 };var res = FindLengthOfLCIS(nums);Console.WriteLine(res);Console.ReadKey();}private static int FindLengthOfLCIS(int[] nums) {//没什么好说的,后面比前面大就计数,用max记录最大的连续的的递增序列if(nums.Length == 0) return 0;int count = 0, max = 0;for(int i = 0; i < nums.Length - 1; i++) {if(nums[i + 1] > nums[i]) {count++;} else {max = Math.Max(max, count);count = 0;}}max = Math.Max(max, count);return max + 1;}}

以上给出1种算法实现,以下是这个案例的输出结果:

该文章的最新版本已迁移至个人博客【比特飞】,单击链接 /archives/3734 访问。

4

分析:

显而易见,以上算法的时间复杂度为: 。

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