LeetCode 413. Arithmetic Slices

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, 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.

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.

解题思路：使用动态规划解决该问题。用dp[i]记录A中从下标0到下标i的连续子集的算数切片个数，用length[i]记录切片结束元素为A[i]的最长切片长度，若无切片，则length[i]=2。那么dp[i]=dp[i-1]+max(length[i]-2,0);关键在于length数组的初始值设置，为什么是2呢？很明显，加入没有规定长度大于等于3时，因为任意两个数都可以组成一个算术切片。代码如下：

class Solution {public:    int numberOfArithmeticSlices(vector<int>& A) {        if(A.empty()||A.size()<3) return 0;        vector<int> dp(A.size(),0),length(A.size(),2);        for(int i = 2; i != A.size(); ++i){            if(A[i]-A[i-1]==A[i-1]-A[i-2]){                length[i]=length[i-1]+1;                dp[i] = dp[i-1]+length[i]-2;            }            else{dp[i] = dp[i-1];}        }        return *(--dp.end());    }};