Arithmetic Slices(详情见leetcode)
1//.题目链接:
https://leetcode.com/problems/arithmetic-slices/description/
https://leetcode.com/problems/arithmetic-slices/description/
2.//题目大意:
给你一个整形一维数组,让你判断长度大于等于3的等差数列又多少个
比如:
Input:
[2, 4, 6, 8, 10]Output: 7
Explanation:
All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]
3.//题目思路:
找规律,长度为n的等差数列,
含有长度至少为3的算数切片的个数为(n-1)(n-2)/2,
在这个数列里,长度为n的有1个,长度为n-1的有2个,则长度为3的有n-2个。。。。总共有(n-1)(n-2)/2个
4//代码
class Solution { public: int numberOfArithmeticSlices(vector<int>& A) { int sumlen=A.size(),result=0,i,len=2;//长度至少要为三 for(i=2;i<sumlen;i++) { if(A[i]-A[i-1]==A[i-1]-A[i-2]) //如果满足等差数列的定义 { len++; } else { if(len>2) {result+=(len-1)*(len-2)*0.5; //对该等差数列进行统计 len=2; } } } if(len>2) result+=(len-1)*(len-2)*0.5; //若是最后是以等差数列形式退出循环,则需要再计时 return result; } };