leetcode-674-Longest Continuous Increasing Subsequence

题目描述：

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. 

 

Note: Length of the array will not exceed 10,000.

 

要完成的函数：

int findLengthOfLCIS(vector<int>& nums) 

 

说明：

1、这道题目给了一个vector，要求找到里面最长的连续升序子vector，返回它的长度，十分容易的一道题。

2、笔者的想法是，从第一位开始找起，迭代下去，直到下一个元素小于等于当前元素，把长度记录下来。

接着从下一个元素开始找起，直到又不满足升序的条件，然后记录长度，与上次得到的长度比较，保留大的数值。

一直处理，直到vector的最后一位。

代码如下：

    int findLengthOfLCIS(vector<int>& nums) 
    {
        int length=1,i=0,s1=nums.size(),max1=1;
        if(s1==0)
            return 0;
        while(i<s1-1)
        {
            if(nums[i]<nums[i+1])
            {
                length++;
                i++;
            }
            else
            {
                max1=max(max1,length);//记录大的长度
                i++;//i更新到下一个元素
                length=1;//重置length
            }
        }
        return max(max1,length);//当碰到类似于[1,3,5,7,9]的测试样例时
    }

上述代码实测16ms，beats 69.47% of cpp submissions。