Longest Continuous Increasing Subsequence（最长递增连续子序列）

题目：

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

给定一个无序的整型数组，找到最长的连续递增子序列。

例1:

输入:[1,3,5,4,7]

输出：3

例2：

输入：[2,2,2,2,2]

输出：1

解法：时间复杂度O(N)

Every (continuous) increasing subsequence is disjoint, and the boundary of each such subsequence occurs whenever nums[i-1] >= nums[i]. When it does, it marks the start of a new increasing subsequence at nums[i], and we store such i in the variable anchor.

For example, if nums = [7, 8, 9, 1, 2, 3], then anchor starts at 0 (nums[anchor] = 7) and gets set again to anchor = 3 (nums[anchor] = 1). Regardless of the value of anchor, we record a candidate answer of i - anchor + 1, the length of the subarray nums[anchor], nums[anchor+1], ..., nums[i]; and our answer gets updated appropriately.

每一个连续递增子序列都是不相交的，并且当nums[i-1]>=nums[i]时，每个子序列的边界产生。我们可以给一个新产生的递增序列的起始点记为nums[i]，然后将此时的i存为一个标记变量a。

比如，数组为[7,8,9,1,2,3]，那么标记变量a为0(nums[a]=7)，然后下一个标记变量a为3(nums[a]=1)。不管标记变量a的值为多少，我们可以将“i-a+1”的值记为子序列nums[a],nums[a+1],...,nums[i]的长度且结果会自适应的更新。

官方代码(Java)：

class Solution {public int findLengthOfLCIS(int[] nums) {int ans = 0, anchor = 0;for (int i = 0; i < nums.length; ++i) {if (i > 0 && nums[i-1] >= nums[i]) anchor = i;ans = Math.max(ans, i - anchor + 1);}return ans;}
}

复杂度分析：

Time Complexity: O(N),where N is the length of nums.

Space Complexity: O(1), the space used by anchor and ans.

时间复杂度：O(N)，N是数组的长度。
空间复杂度：O(1)，空间使用anchor和ans两个变量。

个人代码(C++)：

class Solution {
public:int findLengthOfLCIS(vector<int>& nums) {int l = 1;int ans = l;int len = nums.size();int i = 0;if (len == 0)return 0;for(i=1;i<len;i++){if(nums[i-1]<nums[i]){l++;ans=max(ans,l);}else                l = 1;                }return ans;}
};