LeetCode 674. Longest Continuous Increasing Subsequence(最长连续递增子序列)
题目描述
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.
解法一
思路
这题题目听起来有些误导,可以换成:最长递增子数组。
这题很简单,其实就是顺序遍历数组,然后维持两个变量记录当前最长长度和历史最长长度,然后对每个元素检查其是否满足其比前一个元素值大,若满足则将当前最长长度增加1
;否则更新历史最长长度,并将当前最长长度置为1
(因为当前元素作为新的子序列的第一个元素)。
Python
class Solution:
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 0:
return 0
count = 1
result = 1
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
count += 1
else:
result = max(result, count)
count = 1
result = max(result, count)
return result
Java
class Solution {
public int findLengthOfLCIS(int[] nums) {
if (nums.length == 0) {
return 0;
}
int count = 1;
int result = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
count++;
} else {
result = Math.max(result, count);
count = 1;
}
}
result = Math.max(result, count);
return result;
}
}
C++
class Solution {
public:
int findLengthOfLCIS(vector<int> &nums) {
if (nums.size() == 0) {
return 0;
}
int count = 1;
int result = 1;
for (int i = 1; i < nums.size(); i++) {
if (nums[i] > nums[i - 1]) {
count++;
} else {
result = max(result, count);
count = 1;
}
}
result = max(result, count);
return result;
}
};