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674. Longest Continuous Increasing Subsequence

时间:2021-07-13 11:57:56      阅读:13      评论:0      收藏:0      [点我收藏+]

Given an unsorted array of integers nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.

A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < rnums[i] < nums[i + 1].

 

Example 1:

Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
4.

Example 2:

Input: nums = [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.

 

Constraints:

  • 1 <= nums.length <= 104
  • -109 <= nums[i] <= 109
class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int cur = 1, res = 1;
        for(int r = 1; r < nums.length; r++) {
            if(nums[r] > nums[r - 1]) {
                cur++;
                res = Math.max(res, cur);
            }
            else cur = 1;
        }
        return res;
    }
}

 

674. Longest Continuous Increasing Subsequence

原文:https://www.cnblogs.com/wentiliangkaihua/p/15004740.html

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