397. Longest Continuous Increasing Subsequence

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本文探讨了如何在整数数组中寻找最长的连续递增子序列，提供了一种遍历数组并维护最大递增和递减计数的高效算法，通过两个计数器分别跟踪递增和递减序列长度，最终返回两者之间的最大值。

Longest Continuous Increasing Subsequence

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.
Example
Example 1:

Input: [5, 4, 2, 1, 3]
Output: 4
Explanation:
For [5, 4, 2, 1, 3], the LICS is [5, 4, 2, 1], return 4.
Example 2:

Input: [5, 1, 2, 3, 4]
Output: 4
Explanation:
For [5, 1, 2, 3, 4], the LICS is [1, 2, 3, 4], return 4.
Challenge
O(n) time and O(1) extra space.

解法1：
我的方法是遍历数组，遇到上升序列就累加IncrCount并维护maxIncrCount, 同时DecrCount变成1。同样，遇到下降序列就累加DecrCount并维护maxDecrCount，同时IncrCount变成1。
注意是变成1而不是变成0，因为一个数也算一个序列。
注意，这题跟LintCode 76 Longest Increasing Subsequence那题不一样，LintCode 76 是经典的LIS，只能用DP或单调队列。

class Solution {
public:
    /**
     * @param A: An array of Integer
     * @return: an integer
     */
    int longestIncreasingContinuousSubsequence(vector<int> &A) {
        int len = A.size();
        if (len == 0) return 0;
        int incrCount = 1, maxIncrCount = 1;
        int decrCount = 1, maxDecrCount = 1;
        
        for (int i = 1; i < len; ++i) {
            if (A[i] > A[i - 1]) {
                decrCount = 1;
                incrCount++;
                maxIncrCount = max(maxIncrCount, incrCount);
            } else if (A[i] < A[i - 1]){
                incrCount = 1;
                decrCount++;
                maxDecrCount = max(maxDecrCount, decrCount);
            } else {
                incrCount = 1;
                decrCount = 1;
            }
        }
        
        return max(maxIncrCount, maxDecrCount);
    }
};

解法2： DP?