674. longest continuous increasing sequence

Address ：

Power button https://leetcode-cn.com/problems/longest-continuous-increasing-subsequence/

subject ：

Given an unordered array of integers , Find the longest and Successive increasing subsequences , And return the length of the sequence .

Successive increasing subsequences It can be made up of two subscripts l and r（l < r） determine , If for each l <= i < r, There are nums[i] < nums[i + 1] , So the subsequence [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] It's a continuous increasing subsequence .

Example 1：

Input ：nums = [1,3,5,4,7]
Output ：3
explain ： The longest continuous increasing sequence is [1,3,5], The length is 3.
Even though [1,3,5,7] It's also a subsequence of ascending order , But it's not continuous , because 5 and 7 In the original array is 4 separate . 

Example 2：

Input ：nums = [2,2,2,2,2]
Output ：1
explain ： The longest continuous increasing sequence is [2], The length is 1.
 

Tips ：

1 <= nums.length <= 104
-109 <= nums[i] <= 109

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/longest-continuous-increasing-subsequence
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas ：

Is to compare one by one , Record the maximum length

int findLengthOfLCIS(int* nums, int numsSize){
	int i;
	int idx = 0;
	int maxlen = 1;
	
	for(i=1; i<numsSize; i++)
	{	
		if(nums[i] <= nums[i-1])
		{
			idx = i;
		}
		
		maxlen = fmax(maxlen, i-idx+1);
	}
	
	return maxlen;
}

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