LeetCoce 413. Arithmetic Slices-优快云博客

本文深入探讨了算术切片的概念，详细解释了如何在数组中寻找算术切片的有效算法。通过两种方法——暴力搜索和动态规划，阐述了算法的实现过程，并附有代码示例。此外，还分析了算法的时间复杂度，为读者提供了理解和解决类似问题的思路。

A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

For example, these are arithmetic sequence:

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic.

1, 1, 2, 5, 7

A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.

A slice (P, Q) of array A is called arithmetic if the sequence:
A[P], A[p + 1], …, A[Q – 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.

The function should return the number of arithmetic slices in the array A.

Example:

A = [1, 2, 3, 4]

return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.

代码一

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        if(A.size()<=2) return 0;
        int cnt=0;
        for(int i=0;i<=A.size()-3;i++){
            int temp=A[i+1]-A[i];
            for(int j=i+2;j<A.size();j++)
                if(A[j]-A[j-1]==temp)
                    cnt++;
                else 
                    break;
        }
        return cnt;
    }
};

代码二：动态规划

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int n = A.size();
        if (n < 3) return 0;
        vector<int> dp(n, 0); // dp[i] means the number of arithmetic slices ending with A[i]
        if (A[2]-A[1] == A[1]-A[0]) dp[2] = 1; // if the first three numbers are arithmetic or not
        int result = dp[2];
        for (int i = 3; i < n; ++i) {
            // if A[i-2], A[i-1], A[i] are arithmetic, then the number of arithmetic slices ending with A[i] (dp[i])
            // equals to:
            //      the number of arithmetic slices ending with A[i-1] (dp[i-1], all these arithmetic slices appending A[i] are also arithmetic)
            //      +
            //      A[i-2], A[i-1], A[i] (a brand new arithmetic slice)
            // it is how dp[i] = dp[i-1] + 1 comes
            if (A[i]-A[i-1] == A[i-1]-A[i-2]) 
                dp[i] = dp[i-1] + 1;
            result += dp[i]; // accumulate all valid slices
        }
        return result;
    }
};

转载于:https://www.cnblogs.com/A-Little-Nut/p/10066893.html