Formatted question description: https://leetcode.ca/all/674.html

674. Longest Continuous Increasing Subsequence (Easy)

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.

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Solution 1.

// OJ: https://leetcode.com/problems/longest-continuous-increasing-subsequence/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int ans = 0, len = 0, prev = INT_MAX;
        for (int n : nums) {
            if (n > prev) ++len;
            else len = 1;
            prev = n;
            ans = max(ans, len);
        }
        return ans;
    }
};

Java

  • class Solution {
        public int findLengthOfLCIS(int[] nums) {
            if (nums == null || nums.length == 0)
                return 0;
            int maxLength = 1;
            int increaseLength = 1;
            int length = nums.length;
            for (int i = 1; i < length; i++) {
                if (nums[i] > nums[i - 1])
                    increaseLength++;
                else {
                    maxLength = Math.max(maxLength, increaseLength);
                    increaseLength = 1;
                }
            }
            maxLength = Math.max(maxLength, increaseLength);
            return maxLength;
        }
    }
    
  • // OJ: https://leetcode.com/problems/longest-continuous-increasing-subsequence/
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        int findLengthOfLCIS(vector<int>& nums) {
            int ans = 0, len = 0, prev = INT_MAX;
            for (int n : nums) {
                if (n > prev) ++len;
                else len = 1;
                prev = n;
                ans = max(ans, len);
            }
            return ans;
        }
    };
    
  • class Solution(object):
        def findLengthOfLCIS(self, nums):
            """
            :type nums: List[int]
            :rtype: int
            """
            if not nums: return 0
            N = len(nums)
            dp = [1] * N
            for i in range(1, N):
                if nums[i] > nums[i - 1]:
                    dp[i] = dp[i - 1] + 1
            return max(dp)
    

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