674. Longest Continuous Increasing Subsequence

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本文介绍了一种算法，用于在未排序整数数组中查找最长连续递增子序列的长度。通过遍历数组并使用计数器来跟踪当前递增子序列的长度，当遇到不满足递增条件的元素时，计数器重置为1，并在每次迭代中更新已知的最大长度。

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

Example 1:

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.

Example 2:

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

思路：

1，空数组返回0

2，从头到尾遍历比较，若是满足递增，计算器+1，否则计数器=1，当计数器大于先前保存的最大的数时，更新最大的数

#pragma once
#include<iostream>
#include<vector>

using namespace std;

class Solution {
public:
	int findLengthOfLCIS(vector<int>& nums) {
		//数组为空
		if (nums.size() == 0)
		{
			return 0;
		}

		int maxlen = 1;//最长的
		int t = 1;//计数
		for (int i = 0; i < nums.size() - 1; i++)
		{
			//比较
			if (nums[i] < nums[i + 1])
			{
				t++;
			}
			else {
				t = 1;
			}
			//更新
			if (t > maxlen)
			{
				maxlen = t;
			}
		}

		return maxlen;
	}
};