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committedMar 9, 2025
Add solution #674
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‎README.md

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671|[Second Minimum Node In a Binary Tree](./0671-second-minimum-node-in-a-binary-tree.js)|Easy|
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672|[Bulb Switcher II](./0672-bulb-switcher-ii.js)|Medium|
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673|[Number of Longest Increasing Subsequence](./0673-number-of-longest-increasing-subsequence.js)|Medium|
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674|[Longest Continuous Increasing Subsequence](./0674-longest-continuous-increasing-subsequence.js)|Easy|
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680|[Valid Palindrome II](./0680-valid-palindrome-ii.js)|Easy|
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684|[Redundant Connection](./0684-redundant-connection.js)|Medium|
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686|[Repeated String Match](./0686-repeated-string-match.js)|Easy|
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/**
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* 674. Longest Continuous Increasing Subsequence
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* https://leetcode.com/problems/longest-continuous-increasing-subsequence/
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* Difficulty: Easy
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*
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* Given an unsorted array of integers nums, return the length of the longest continuous increasing
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* subsequence (i.e. subarray). The subsequence must be strictly increasing.
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*
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* A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is
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* [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].
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*/
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var findLengthOfLCIS = function(nums) {
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let result = 1;
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for (let i = 1, v = 1; i < nums.length; i++) {
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v = nums[i] > nums[i-1] ? v + 1 : 1;
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result = Math.max(result, v);
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}
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return result;
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};

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