leetcode 674. Longest Continuous Increasing Subsequence（最长连续递增子序列）

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
4.

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.

找出最长的连续递增子序列，注意是连续的，而且是绝对递增的

思路：
遍历数组，如果满足绝对大于，count就加1，否则count重置为1

    public int findLengthOfLCIS(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        
        int count = 1;
        int result = 1;
        int n = nums.length;
        
        for(int i = 1; i < n; i++) {
            if(nums[i] > nums[i-1]) {
                count ++;
                result = Math.max(result, count);
            } else {
                count = 1;
            }
        }
        
        return result;
    }