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

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.

这题挺好的，我想到两种方法，一种brute force，O(n2)不好；另一种，看到LIS就思维定势地想到了DP。

DP:

我写的leetcode submissions没保存，摘抄一个别人的：

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

但其实这题有更好的方法，

public int findLengthOfLCIS(int[] nums) {        int res = 0, cnt = 0;        for(int i = 0; i < nums.length; i++){            if(i == 0 || nums[i-1] < nums[i]) res = Math.max(res, ++cnt);            else cnt = 1;        }        return res;    }复制代码

用O(1)记录某个值然后随着滚动而重置，这种思想。