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:
- 使用貪心算法的思想找出更長(zhǎng)的連續(xù)遞增序列
- 記錄該序列的起點(diǎn)下標(biāo)，終點(diǎn)下標(biāo)，即可輕而易舉求出長(zhǎng)度
- 最后，每次循環(huán)中取長(zhǎng)度的最大值。
AC文章來(lái)源：http://www.zghlxwxcb.cn/news/detail-467572.html
/*
* @lc app=leetcode.cn id=674 lang=cpp
*
* [674] 最長(zhǎng)連續(xù)遞增序列
*/
// @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
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