[LeetCode] 674. Longest Continuous Increasing Subsequence

轉自LeetCode

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

Example 1:

Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. 
Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4. 

Example 2:

Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1. 

Note: Length of the array will not exceed 10,000.

<Solution>

想法如下

• 用 slicing window 的想法。當 nums[i-1] >= nums[i] 時，將左邊界移到 i，重新開始找

• 每次 iteration 時，都去計算當下的最長長度

code 如下

Java

	class Solution {
	public int findLengthOfLCIS(int[] nums) {
	if(nums.length < 2) {
	return nums.length == 0 ? 0 : 1;
	}
	int anchor = 0;
	int ans = Integer.MIN_VALUE;
	for(int i = 0; i < nums.length; i++) {
	if(i > 0 && nums[i-1] >= nums[i]) {
	anchor = i;
	}
	ans = Math.max(ans, i - anchor + 1);
	}
	return ans;
	}
	}

view raw findLengthOfLCIS.java hosted with ❤ by GitHub

Kotlin

	class Solution {
	fun findLengthOfLCIS(nums: IntArray): Int {
	var left = 0
	var ans = Int.MIN_VALUE
	for(right in nums.indices) {
	if(right > 0 && nums[right - 1] >= nums[right] ) {
	left = right
	}
	ans = Math.max(ans, right - left + 1)
	}
	return ans
	}
	}

view raw longest_continuous_increasing_subsequence.kt hosted with ❤ by GitHub