0674. Longest Continuous Increasing Subsequence

# 674. Longest Continuous Increasing Subsequence#

## 题目 #

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.

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  with length 1. Note that it must be strictly
increasing.
``````

Constraints:

• `0 <= nums.length <= 10^4`
• `10^9 <= nums[i] <= 10^9`

## 解题思路 #

• 简单题。这一题和第 128 题有区别。这一题要求子序列必须是连续下标，所以变简单了。扫描一遍数组，记下连续递增序列的长度，动态维护这个最大值，最后输出即可。

## 代码 #

``````package leetcode

func findLengthOfLCIS(nums []int) int {
if len(nums) == 0 {
return 0
}
res, length := 1, 1
for i := 1; i < len(nums); i++ {
if nums[i] > nums[i-1] {
length++
} else {
res = max(res, length)
length = 1
}
}
return max(res, length)
}

func max(a, b int) int {
if a > b {
return a
}
return b
}
`````` Nov 25, 2022 Edit this page