Leetcode_Array -- 674. Longest Continuous Increasing Subsequence [easy]

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

给定一个未排序的整数数组，找到最长的连续递增子序列，返回其长度

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.

Solutions：

Python

'''
方法1是leetcode最快算法，算法2是自己想到的算法，其实思路都一样，自己想的麻烦了一些，可以用方法1优化
算法的主要思想就是比较当前数是否小于下一个数，如果小于则进行计数，count++，并同时更新maxcount,
否则count置1，最终返回maxcount，由于maxcount最小是1，所以还需要考虑nums为空的情形，单独返回0即可
'''

(1)
class Solution:
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        maxLen = 1
        curLen = 1
        for i in range(1,len(nums)):
            if nums[i] > nums[i-1]:
                curLen += 1
                if curLen > maxLen:
                    maxLen = curLen 
            else:
                curLen = 1
        return maxLen

(2)
'''
感觉自己改进上面的代码应该运行速度比上面更快，事实却并没有,
leetcode的计时有的时候也有些问题，
总之算法思路完全一样
'''
class Solution:
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        maxLen = 1
        curLen = 1
        for i in range(1,len(nums)):
            if nums[i] > nums[i-1]:
                curLen += 1
            else:
                if curLen > maxLen:
                    maxLen = curLen 
                curLen = 1
        if curLen > maxLen:
            maxLen = curLen
        return maxLen

(3)
class Solution:
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) ==0:
            return 0
        start = 0
        end = 0
        length = 0
        temp = nums[0]
        
        for i in range(1,len(nums)):
            if nums[i] > temp:
                end = i
                temp = nums[i]
            else:
                if end-start + 1 > length:
                    length = end - start + 1
                start = i 
                temp = nums[i]
                
        if end - start +1 > length:
            length = end - start + 1
        return length

C++

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        if (nums.size() == 0){
            return 0;
        }
        int curlen = 1;
        int maxlen = 1;
        for (int i = 1;i<nums.size();i++){
            if (nums[i]>nums[i-1]){
                curlen ++;
            }
            else{
                if (curlen>maxlen){
                    maxlen = curlen;
                }
                curlen = 1;
            }
        }
        if (curlen>maxlen){
            maxlen = curlen;
        }
        return maxlen;
    }
};