Week10

Problem–Medium–413. Arithmetic Slices

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.

题目解析：

由于我的代码状态转移方法不好，所以下面根据答案进行解析。 这是一道动态规划的题目，由题意我们很容易将问题分解为更小的子问题，即原来求长度为n的数组的算术分割有多少个，我们可以先求长度更小的数组如n-1时算术分割有多少个，然后进行状态转移。具体实现就是将dp[n]看成以n结尾算术分割个数可以通过求以n-1结尾即dp[n-1]来求得，然后将所有结果累加即可。 假设长度为n的数组都是等差的，那么dp[n]表示长度以n结尾的算术分割的个数。因为dp[3] = 1, dp[4] = 2, dp[5] = 3…如果以n-1结尾有算术分割，那么以n结尾的算术分割也包括n-1的算术分割个数，在n-1算术分割的基础上加一即可。状态转移方程为dp[n] = dp[n-1] + 1 (A[n] – A[n-1] == A[n-1] – A[n – 2])，而所求的结果就是他们的累加之和。

代码：

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int n = A.size();
        if (n < 3) return 0;
        vector<int> dp(n, 0);
        if (A[2]-A[1] == A[1]-A[0]) dp[2] = 1;
        int result = dp[2];
        for (int i = 3; i < n; ++i) {
            if (A[i]-A[i-1] == A[i-1]-A[i-2]) 
                dp[i] = dp[i-1] + 1;
            result += dp[i];
        }
        return result;
    }
};