Longest Continuous Increasing Subsequence

最新推荐文章于 2024-09-28 18:22:46 发布

转载最新推荐文章于 2024-09-28 18:22:46 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/jiadyang/p/8612827.html

本文介绍了如何寻找未排序整数数组中最长连续递增子序列的长度，并提供了两种解决方案，一种是非动态规划方法，另一种是使用动态规划方法。

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.

讲道理这题最先想到用DP,不过还是先上一个非DP的吧：）

 1 class Solution {
 2 public:
 3     int findLengthOfLCIS(vector<int>& nums) {
 4         
 5         if(nums.empty())
 6             return 0;
 7         
 8         int n = nums.size();
 9         
10         int cnt = 1, res = 0;
11         
12         for(int i = 1; i < n;++i)
13         {
14             if(nums[i] > nums[i - 1])
15             {
16                 ++cnt;
17             }
18             else
19             {
20                 res = max(res, cnt);
21                 cnt = 1;
22             }
23         }
24         
25         return max(res, cnt);
26     }
27 };

再上个DP的：

 1 class Solution {
 2 public:
 3     int findLengthOfLCIS(vector<int>& nums) {
 4         
 5         if(nums.empty())
 6             return 0;
 7         
 8         int n = nums.size();
 9         int res = 1;
10         vector<int> dp(n, 1);
11         
12         for(int i = 1; i < n;++i)
13         {
14             if(nums[i] > nums[i - 1])
15             {
16                 dp[i] = dp[i-1] + 1;
17             }
18             else
19             {
20                 dp[i] = 1;    
21             }
22             
23             res = max(res, dp[i]);
24         }
25         
26         return res;
27     }
28 };