0674.LongestContinuousIncreasingSubsequence.js

// 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

// 来源：力扣（LeetCode）
// 链接：https://leetcode-cn.com/problems/longest-continuous-increasing-subsequence
// 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

// 🎨 方法一：动态规划


/**
 * @param {number[]} nums
 * @return {number}
 */
var findLengthOfLCIS = function (nums) {
    let n = nums.length;

    if (n === 0) {
        return 0
    }

    const dp = new Array(n).fill(1);

    let res = 1;
    for (let i = 1; i < n; i++) {
        if (nums[i] > nums[i - 1]) {
            dp[i] = dp[i - 1] + 1
            res = Math.max(dp[i], res)
        }
    }

    return res
};