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
-
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
Thought:
- 使用贪心算法的思想找出更长的连续递增序列
- 记录该序列的起点下标,终点下标,即可轻而易举求出长度
- 最后,每次循环中取长度的最大值。
AC
/*
* @lc app=leetcode.cn id=674 lang=cpp
*
* [674] 最长连续递增序列
*/
// @lc code=start
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int ans = 0;
int n = nums.size();
int start = 0;
for(int i = 0; i < n; i++)
{
if(i > 0 && nums[i] <= nums[i - 1])
{
start = i;
}
ans = max(ans, i - start + 1);
}
return ans;
}
};
// @lc code=end