Leetcode编程练习：Arithmetic Slices

题目原文：（id=413)

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.

给定一个数列，找到其中为等差数列的子串的个数。方法是从前往后对数列进行扫描，每次把当前元素加入先前考虑的数列中，判断增加了多少为等差数列的子串。显然这些子串肯定包含最后一个元素，只要记录上次迭代时倒数2个元素的差值d，再判断现在倒数2个元素差值是否与记录的数字d相同，是则增加的子串数为之前差值为d的子串数再加1.扫描一遍数字时间复杂度为O(n)

	int numberOfArithmeticSlices(vector& A) {
		int ans1, ans2;
		ans1 = 0;
		ans2 = 0;
		int count=0, d;
		if (A.size()>1) d=A[1]-A[0];
		for (int i = 2;i < A.size();i++) {
			if (d == A[i] - A[i - 1]) {
				count = count+1;
			}
			else {
				count = 0;
				d = A[i] - A[i - 1];
			}
			if (i % 2) 
				ans1 = ans2 + count;
			else
				ans2 = ans1 + count;
		}
		return max(ans1, ans2);
	}