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.
題意&解題思路
arithmetic slices的定義是長度至少爲3的數組段，其段間差值相等，要求所給數組一共會有多少個arithmetic slices。
如：1,2,3,4,5 這樣的數組可以切出
[1,2,3] [2,3,4] [3,4,5] [1,2,3,4] [2,3,4,5] [1,2,3,4,5]
這6種情況。
用動態規劃來處理該題目
我們用dp[i]來記錄以A[i]結束的數組中有多少種arithmetic slices，則
若A[i] - A[i - 1] == A[i - 1] - A[i - 2], dp[i] = dp[i - 1] + 1
因爲當前值加在A[i - 1]後可以形成一個新的arithmetic slices，另外，以A[i - 1]結尾的最長slices數組s內含有的slices總數爲dp[i - 1]，則去掉s的頭一個元素，再在A[i-1]後加入A[i]，其形成的新數組s’和s的結構是一樣的，故s’的slices數量等於dp[i - 1]。
由這個關係，我們再將各個位置上的dp值相加起來，即可得到答案。
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& A) {
int num = A.size();
if(num < 3)return 0;
int dp[num + 1] = {0};
if(A[1] - A[0] == A[2] - A[1])dp[2] = 1;
int ans = dp[2];
for(int i = 3; i < num; i++){
if(A[i] - A[i - 1] == A[i - 1] - A[i - 2])dp[i] = dp[i - 1] + 1;
ans += dp[i];
}
return ans;
}
};