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— Leetcode

preface

Hello, everyone. I’m One.

Confused algorithm, rarely confused

Question

674. Longest continuously increasing sequence

Difficulty: Easy

Given an unsorted array of integers, find the longest continuously increasing subsequence and return the length of that sequence.

Continuously increasing subsequences can be determined by two subscripts L and r (l < r), if for every L <= I < r, nums[I] < nums[I + 1], Then the subsequence [NUMs [L], NUMs [L + 1]…, NUMs [r-1], NUMs [r]] is a continuous increasing subsequence.

Example 1:

Input: nums = [1,3,5,4,7] output: 3 explanation: the longest continuously increasing sequence is [1,3,5] and has a length of 3.Copy the code

Although [1,3,5,7] is also an ascending subsequence, it is not continuous because 5 and 7 are separated by 4 in the original array. Example 2:

Input: nums = [2,2,2,2] Output: 1 Explanation: The longest continuously increasing sequence is [2], of length 1.Copy the code

Solution

The same thing as the largest number of consecutive ones

The maximum number of consecutive 1s

Code

All LeetCode codes have been synchronized to Github

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/ * * *@authorA coding * /
class Solution {
    public int findLengthOfLCIS(int[] nums) {
        if(nums.length <= 1)
            return nums.length;
        int ans = 1;
        int count = 1;
        for(int i=0; i<nums.length-1; i++) {if(nums[i+1] > nums[i]) {
                count++;
            } else {  
                count = 1;
            }
            ans = count > ans ? count : ans;
        }
        returnans; }}Copy the code

Result

Complexity analysis

  • Time complexity: O(N)

Last

One foot is difficult to go, one hand is difficult to sing, a person’s power is limited after all, a person’s journey is doomed to be lonely. When you set a good plan, with full of blood ready to set out, must find a partner, and Tang’s monk to the West, the master and his disciples four people united as one to pass the ninety-eight difficult. So,

If you want to get into a big factory,

To learn data structures and algorithms,

If you want to keep brushing,

To have a group of like-minded people,

Please join the team to brush the questions