[leetcode 413. Arithmetic Slices]medium|week 12

一、题目

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.
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二、思路分析

这道题我一开始想这用动态数组的思想来做，后面发现了我其实对题目理解有偏差，我以为找的等差数列不需要连起来，然后怎样都想不出来要怎么做。后面发现原来题目中的等差数列必须是连起来的，那就比较好办了，用数学的方法就可以得到答案。因为数字数量为n的等差数列有n(n+1)/2个子列，所以每当遍历一个数量大于等于3的等差数列，就设置一个变量用来对每个等差数列进行累加。最后即可得到答案。

三、代码

#include<vector>using namespace std;class Solution {public:    int numberOfArithmeticSlices(vector<int>& A) {        int count = 0;        int num = 0;        for (int i=2;i<A.size();i++){             if (A[i]-A[i-1]==A[i-1]-A[i-2]){                 num++;        //对于一个数量为n的等差数列，子列数量为 n(n+1)/2;                   count+=num;                 }             else                 num=0;    //当无法构成等差时递增值设为0         }         return count;    }    };