674. Longest Continuous Increasing Subsequence.md

674. Longest Continuous Increasing Subsequence

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leetcode-cn Daily Challenge on January 24th, 2021.

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Difficulty : Easy

Related Topics : Array

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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

• mine Your runtime beats 99.95 % of java submissions.

// O(n)time O(1)space
class Solution {
    public int findLengthOfLCIS(int[] nums) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        if(nums.length == 1){
            return 1;
        }
        int max = 1, temp = 1;
        for(int i = 1; i < nums.length; i++){
            if(nums[i] > nums[i - 1]){
                temp++;
                max = Math.max(temp, max);
            }else{
                temp = 1;       
            }
        }
        return max;
    }
}

• the most votes Your runtime beats 99.95 % of java submissions.

class Solution {
    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;
    }
}