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.

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:

• 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
}