题目

题目挺简单：找一个最长递增子串

我的代码

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        
        if not nums:
            return 0
        i,result_li,row = 0,[],[nums[0]]
        
        while i<len(nums)-1:
            if nums[i]<nums[i+1]:
                  row.append(nums[i+1])
            else:
                result_li.append(len(row))
                row=[nums[i+1]]
            i+=1
        result_li.append(len(row))
        return max(result_li)

优秀代码
思路与我差不多，就是不存储信息，直接判断是否大于。

class Solution:
    def findLengthOfLCIS(self, nums: 'List[int]') -> 'int':
        """
        1,3,5,4,7 -> 3
        2,2,2,2,2 -> 1
        """
        if not nums:
            return 0
        
        curr_len = 1
        max_len = 0
        prev = nums[0]
        for num in nums[1:]:
            # print(f'num: {num}, prev: {prev}, curr_len: {curr_len}')
            if num > prev:
                curr_len += 1
            else:
                max_len = max(max_len, curr_len)
                curr_len = 1
            prev = num
        
        max_len = max(max_len, curr_len)
        return max_len