LeetCode 674 - Longest Continuous Increasing Subsequence

https://leetcode.com/problems/longest-continuous-increasing-subsequence/

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

https://leetcode.com/articles/longest-continuous-increasing-subsequence/

  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;

  }

https://leetcode.com/problems/longest-continuous-increasing-subsequence/discuss/107352/Java-code-6-liner

public int findLengthOfLCIS(int[] nums) {
        if(nums.length==0) return 0;
        int length=1,temp=1;
        for(int i=0; i<nums.length-1;i++) {
            if(nums[i]<nums[i+1]) {temp++; length=Math.max(length,temp);}
            else temp=1; 
        }
        return length;
    }

follow up1 要求subsequence中前后两个元素最⼤大间隔为1
https://github.com/mintycc/OnlineJudge-Solutions/blob/master/untag/LCIS%20Gap%20One/LCISGapOne.java

    public int findLengthOfLCIS(int[] nums) {
        if (nums.length < 2) return nums.length;
        int pp = 1;
        int p = nums[0] < nums[1] ? 2 : 1;
        int ans = p;
        
        for (int i = 2; i < nums.length; i ++) {
            int tmp = 1;
            if (nums[i - 2] < nums[i] && pp + 1 > tmp) tmp = pp + 1;
            if (nums[i - 1] < nums[i] && p  + 1 > tmp) tmp = p  + 1;
            ans = Math.max(ans, tmp);
            pp = p; p = tmp;
        }
        
        return ans;
    }

follow up2 要求subsequence中前后两个元素最⼤大间隔为k
https://github.com/mintycc/OnlineJudge-Solutions/blob/master/untag/LCIS%20Gap%20One/LCISGapK.java

    public int findLengthOfLCIS(int[] nums, int k) {
        if (nums.length == 0) return 0;
        
        int[] f = new int[nums.length];
        Arrays.fill(f, 1);      // every number itself can be a shortest increasing subsequence
        
        int ans = 1;
        for (int i = 1; i < nums.length; i ++) {
            for (int j = Math.max(0, i - k - 1); j < i; j ++)   // the limit of gap k
                if (nums[i] > nums[j])                          // make sure subsequence is increasing
                    f[i] = Math.max(f[i], f[j] + 1);
            ans = Math.max(ans, f[i]);                  // update result
        }
        return ans;
    }

follow up3 允许⼀一个break Example:
Input: [7, 3, 2, 3, 5, 6, 4, 2, 1]
Output: 5 - [2, 3, 5, 6, 4], [6, 4] is the break OR [3, 2, 3, 5, 6], [3,2] is the break
要有⼀一个变量量判断当前状态是否break过, 还有breaking index是另⼀一个 subarray的开始