[LeetCode] 674. Longest Continuous Increasing Subsequence 最长连续递增序列

2021年09月15日 阅读数:1
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Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).html

Example 1:java

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:python

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.数组

给一个没有排序的整数数组,找出最长的连续递增子序列(子数组)。post

Java:url

public int findLengthOfLCIS(int[] nums) {
        int res = 0, cnt = 0;
        for(int i = 0; i < nums.length; i++){
            if(i == 0 || nums[i-1] < nums[i]) res = Math.max(res, ++cnt);
            else cnt = 1;
        }
        return res;
    }  

Python:code

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result, count = 0, 0
        for i in xrange(len(nums)):
            if i == 0 or nums[i-1] < nums[i]:
                count += 1
                result = max(result, count)
            else:
                count = 1
        return result  

Python: wohtm

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        
        n = len(nums)
        dp = [0] * n
        dp[0] = 1
        longest = 1
        for i in xrange(1, n):
            if nums[i] > nums[i-1]:
                dp[i] = dp[i-1] + 1
            else:
                dp[i] = 1
            longest = max(longest, dp[i])
  
        return longest  

C++:blog

int findLengthOfLCIS(vector<int>& nums) {
        int res = 0, cnt = 0;
        for(int i = 0; i < nums.size(); i++){
            if(i == 0 || nums[i-1] < nums[i]) res = max(res, ++cnt);
            else cnt = 1;
        }
        return res;
    }

 

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