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 < 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:

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

题目大意 #

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。

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
}

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