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LeetCode 413. Arithmetic Slices

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题目

原题在此

解析

  1. 3个能构成1个Slices, 4个能构成1+2个, 5个能构成1+2+3个
  2. 每次比较三个就行, count每次比较成功就+1, 失败就归零.

代码

c++

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