Arithmetic Slices动态规划算法详解

问题详见： Arithmetic Slices

题目让我们求解一个给定的数组中等差子序列的个数。题目描述如下：

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.

解题思路：

由于题目中等差子序列的长度至少为3，所以如果所给数组长度小于3那么结果即为0，否则就进行动态规划。动态规划的思路是记录以当前遍历到的数组元素为子序列尾部元素的等差子序列的个数dp[i]，而dp[i]=dp[i−1]+1。这样遍历完以后将dp[2]到dp[n−1]加起来即为所求答案。整个算法复杂度为O(n)。具体算法如下：

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;
}
};

其提交运行结果如下：