674-最长连续递增序列(Longest Continuous Increasing Subsequence)

原文链接: https://leetcode-cn.com/problems/longest-continuous-increasing-subsequence

英文原文

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 < r, nums[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

中文题目

给定一个未经排序的整数数组，找到最长且 连续递增的子序列，并返回该序列的长度。

连续递增的子序列 可以由两个下标 l 和 r（l < r）确定，如果对于每个 l <= i < r，都有 nums[i] < nums[i + 1] ，那么子序列 [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] 就是连续递增子序列。

示例 1：

输入：nums = [1,3,5,4,7]
输出：3
解释：最长连续递增序列是 [1,3,5], 长度为3。
尽管 [1,3,5,7] 也是升序的子序列, 但它不是连续的，因为 5 和 7 在原数组里被 4 隔开。

示例 2：

输入：nums = [2,2,2,2,2]
输出：1
解释：最长连续递增序列是 [2], 长度为1。

提示：

1 <= nums.length <= 10⁴
-10⁹ <= nums[i] <= 10⁹

通过代码

高赞题解

解题思路

标签：遍历
过程：
- count 为当前元素峰值，ans为最大峰值
- 初始化 count = 1
- 从 0 位置开始遍历，遍历时根据前后元素状态判断是否递增，递增则 count++，递减则 count=1
- 如果 count>ans，则更新 ans
- 直到循环结束
时间复杂度：$O(N)$

代码

[]
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;
        }
        return ans;
    }
}