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

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

问题

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

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:

4

分析:

显而易见,以上算法的时间复杂度为: O(n) 。

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