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

Example 1:
 Input:
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:
Output: 1
 Explanation: The longest continuous increasing subsequence is [2], its length is 1.
Note: Length of the array will not exceed 10,000. 

int findLengthOfLCIS(int* nums, int numsSize) 
{
    if(NULL == nums || 0 == numsSize )
        return 0;
    
    /*用于统计最大连续子数组长度*/
    int max = 1;
    /*cnt是统计的当前连续子数组长度*/
    int cnt = 1;
    
    for(int i = 1;i < numsSize;i++)
    {
        if(nums[i] > nums[i-1])
        {
            /*cnt初始值为1，因为存在大小关系至少需要两个，后续每存在一个大的，加1*/
            cnt += 1;
        }
        else
        {
            max = (max > cnt) ? max:cnt;
            /*重置cnt，开始新一轮统计*/
            cnt = 1;
        }
        
    }
    /*如果数组所有元素均递增，不进入else分支，在此添加判断*/
    max = (max > cnt) ? max:cnt;
    
    return max;
    
}