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.

讲道理这题最先想到用DP,不过还是先上一个非DP的吧：）

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

再上个DP的：

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