leetcode 413. Arithmetic Slices

1.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.
等差数列的定义为：至少三个数字组成，两两相邻的数字之差相同。
等差数列，例如：
1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9
不是等差数列：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.
一个下标从0开始的数组A，长度为N。 一个数组切片(P,Q), 满足0<=P

2.分析

两种方法。
方法1： solution里看到的dp
设dp[i]表示以A[i]为结尾的切片数量，那么dp[i+1]=dp[i]+1
例如 1 1 2 3 4
以3为结尾的切片有 1 2 3
以4位结尾的切片有 1 2 3 4 ， 2 3 4 多出来的就是2 3 4，所以dp[i+1]=dp[i]+1
按照这个递推公式依次求解，求和。
方法2：找出每一段等差数列，根据长度求可能的切片个数。
例如： 1 1 2 3 4
1 2 3 4是等差数列，长度为切片的可能有1 2 3、2 3 4，1 2 3 4
长度为x的等差数列切片的数量为 x-2+x-3+x-4…+1=(x-2)(x-1)/2
最后求总和即可。

3.代码

方法1：

class Solution {public:    int numberOfArithmeticSlices(vector<int>& A) {        int count = 0;        int total = 0;        int j = 2;        while (j < A.size()) {            if (A[j-1] - A[j - 2] == A[j] - A[j-1])                count += 1;            else                count = 0;            total += count;            ++j;        }        return total;    }};

方法2

int numberOfArithmeticSlices(vector<int>& A) {    if (A.size() < 3)        return 0;    int* diff = new int[A.size() - 1];    for (int i = 1; i < A.size(); i++)        diff[i - 1] = A[i] - A[i - 1];//相邻元素的差    int total = 0;    int i = 0;    while (i < A.size() - 1){        int j = i;        while (j < A.size() - 2 && diff[j] == diff[j + 1])            ++j;        if (j > i)            total += (j - i)* (j - i + 1) / 2;        i = j + 1;    }    return total;}