674 Longest Continuous Increasing Subsequence 最长连续递增序列
Description:
Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).
Example:
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.
题目描述:
给定一个未经排序的整数数组，找到最长且连续的的递增序列。
示例 :
示例 1:
输入: [1,3,5,4,7]
输出: 3
解释: 最长连续递增序列是 [1,3,5], 长度为3。
尽管 [1,3,5,7] 也是升序的子序列, 但它不是连续的，因为5和7在原数组里被4隔开。
示例 2:
输入: [2,2,2,2,2]
输出: 1
解释: 最长连续递增序列是 [2], 长度为1。
注意：数组长度不会超过10000。
思路:
要求最长连续子数组, 遍历数组, 只要不满足递增条件, 重新从 1开始计数, 返回最大的计数值即可
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int findLengthOfLCIS(vector& nums)
{
int result = nums.size() < 2 ? nums.size() : 0, temp = 1;
for (int i = 1; i < nums.size(); i++)
{
temp = nums[i] > nums[i - 1] ? temp + 1 : 1;
result = max(temp, result);
}
return result;
}
};
Java:
class Solution {
public int findLengthOfLCIS(int[] nums) {
int result = nums.length < 2 ? nums.length : 0, temp = 1;
for (int i = 1; i < nums.length; i++) {
temp = nums[i] > nums[i - 1] ? temp + 1 : 1;
result = Math.max(temp, result);
}
return result;
}
}
Python:
class Solution:
def findLengthOfLCIS(self, nums: List[int]) -> int:
result, temp = 0 if len(nums) > 1 else len(nums), 1
for i in range(1, len(nums)):
temp = temp + 1 if nums[i] > nums[i - 1] else 1
result = max(result, temp)
return result