#397 Longest Increasing Continuous Subsequence

最新推荐文章于 2024-07-22 23:02:20 发布

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最新推荐文章于 2024-07-22 23:02:20 发布

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分类专栏： lintcode 文章标签： lintcode dynamic programming

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本文介绍了一种高效算法，用于寻找整数数组中最长连续递增子序列，包括从左到右或从右到左的增长序列。通过动态规划思想实现，确保O(n)的时间复杂度和O(1)的空间复杂度。

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题目描述：

Give an integer array，find the longest increasing continuous subsequence in this array.

An increasing continuous subsequence:

Can be from right to left or from left to right.
Indices of the integers in the subsequence should be continuous.

Notice

O(n) time and O(1) extra space.

Have you met this question in a real interview?

Yes

Example

For [5, 4, 2, 1, 3], the LICS is [5, 4, 2, 1], return 4.

For [5, 1, 2, 3, 4], the LICS is [1, 2, 3, 4], return 4.

题目思路：

这题用了一点dp的思想：用local来记录小范围内的增长/减少subarray的长度，如果local_inc遇到减少的情况，就reset成1；用global来记录最长的长度，取自己和local的max。

Mycode（AC = 22ms）：

class Solution {
public:
    /**
     * @param A an array of Integer
     * @return  an integer
     */
    int longestIncreasingContinuousSubsequence(vector<int>& A) {
        // Write your code here
        int local_inc = 1, local_dec = 1;
        int global_inc = 1, global_dec = 1;
        
        if (A.size() <= 2) return A.size();
        
        // local_inc to record locally max increasing length
        // local_dec to record locally max decreasing length
        for (int i = 1; i < A.size(); i++) {
            if (A[i] >= A[i - 1]) {
                local_inc++;
                global_inc = max(global_inc, local_inc);
                local_dec = 1;
            }
            else {
                local_dec++;
                global_dec = max(global_dec, local_dec);
                local_inc = 1;
            }
        }
        
        return max(global_inc, global_dec);
    }
};