【Leecode】674. The longest continuous increasing sequence

Given an unsorted array of integers nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.

A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].

Example 1:

Input: nums = [1,3,5,4,7]
Output: 3

Explanation:

• The longest continuous increasing subsequence is [1,3,5] with length 3.
• Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element

Example 2:

Input: nums = [2,2,2,2,2]
Output: 1

Explanation:

• The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly increasing.

Constraints:

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

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Thought:

• Use the idea of ​​​​greedy algorithm to find a longer continuous increasing sequence
• Record the start subscript and end subscript of the sequence, and you can easily calculate the length
• Finally, take the maximum length in each loop.

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AC

/*
 * @lc app=leetcode.cn id=674 lang=cpp
 *
 * [674] 最长连续递增序列
 */

// @lc code=start
class Solution {
    
    
public:
    int findLengthOfLCIS(vector<int>& nums) {
    
    
        int ans = 0;
        int n = nums.size();
        int start = 0;
        for(int i = 0; i < n; i++)
        {
    
    
            if(i > 0 && nums[i] <= nums[i - 1])
            {
    
    
                start = i;
            }
            ans = max(ans, i - start + 1);
        }
        return ans;
    }
};
// @lc code=end

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