Algorithm
Problem Name: 674. Longest Continuous Increasing Subsequence
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
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
if(nums.empty()) return 0;
int i = 0, j = 1, maxlen = 1;
while(j < nums.size()){
while(j < nums.size(> && nums[j] > nums[j - 1]) j++;
maxlen = max(maxlen, j - i);
i = j++;
}
return maxlen;
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int findLengthOfLCIS(int[] nums) {
int maxLength = 1;
int idx = 0;
while (idx < nums.length) {
int count = 1;
while (idx < nums.length && idx + 1 < nums.length && nums[idx] < nums[idx + 1]) {
idx++;
count++;
}
maxLength = Math.max(maxLength, count);
idx++;
}
return maxLength;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const findLengthOfLCIS = function(nums) {
if (nums.length == 1) {
return 1;
}
let ans = 0,
anchor = 0;
for (let i = 1; i < nums.length; i++) {
if (nums[i - 1] >= nums[i]) {
anchor = i;
}
ans = Math.max(ans, i - anchor + 1);
}
return ans;
};
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if nums==[]: return 0
curr, mx=1, 1
for i in range(len(nums)-1):
if nums[i+1]>nums[i]: curr+=1; mx=max(mx,curr)
else: curr=1
return mx
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0674_LongestContinuousIncreasingSubsequence
{
public int FindLengthOfLCIS(int[] nums)
{
if (nums.Length < = 1) return nums.Length;
var max = 1;
var current = 1;
for (int i = 1; i < nums.Length; i++)
{
if (nums[i] > nums[i - 1])
{
current++;
max = Math.Max(max, current);
}
else
current = 1;
}
return max;
}
}
}
Input
Output