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;
    }
};