[LeetCode] Arithmetic Slices 等差数列的个数

声明：原题目转载自LeetCode，解答部分为原创

Problem ：

        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, 97, 7, 7, 73, -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.

Solution :

         思路：用一个数组 differ[] 来记录 A数组相邻两位元素的差值，再将 differ[] 数组按值划分成多个区间，记每个区间的长度为same_num， 则该区间可形成的等差数组共有 same_num * (same_num - 1) / 2 个。将各个值相加可得 A[] 数组可形成的等差数列总个数。

         代码如下：

#include<iostream>#include<vector>using namespace std;class Solution {public:    int numberOfArithmeticSlices(vector<int>& A) {    int size = A.size();    if(size < 3)    return 0;    vector<int> differ(size - 1, 0);    for(int i = 0 ; i < differ.size() ; i ++)    {    differ[i] = A[i + 1] - A[i];}int num = 0;int same_num = 1;for(int i = 1 ; i < differ.size() ; i ++){if(differ[i] == differ[i - 1])same_num ++;else{num += same_num * (same_num - 1) / 2;same_num = 1;}if(i == differ.size() - 1 && same_num > 1){num += same_num * (same_num - 1) / 2;}}return num;}};int main(){vector<int> temp;temp.push_back(1);temp.push_back(2);temp.push_back(3);temp.push_back(8);temp.push_back(9);temp.push_back(10);Solution text;cout << text.numberOfArithmeticSlices(temp) << endl;return 0;}