446. Arithmetic Slices II

题目：A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.A zero-indexed array A consisting of N numbers is given. A subsequence slice of that array is any sequence of integers (P0, P1, ..., Pk) such that 0 ≤ P0 < P1 < ... < Pk < N. A subsequence slice (P0, P1, ..., Pk) of array A is called arithmetic if the sequence A[P0], A[P1], ..., A[Pk-1], A[Pk] is arithmetic. In particular, this means that k ≥ 2. The function should return the number of arithmetic subsequence slices in the array A.

The input contains N integers. Every integer is in the range of -231 and 231-1 and 0 ≤ N ≤ 1000. The output is guaranteed to be less than 231-1.

原题目太长了，大意就是定义了等差数列，是一个长度大于等于3的数列，给定一个数列，找出里面等差数列的个数，等差数列的长度大于等于3就可以，数列可以不连续。注意整数的范围是 -231 到 231-1。给定数列A的长度N范围是0 ≤ N ≤ 1000。

思路：之前做了一道题413 Arthmetic Slices，那道题严格要求等差数列是连续的，于是用一个一维数组dp[n]表示以第n个数字为结尾的等差数列个数。初始化dp全为0。如果A[2]-A[1]=A[1]-A[0]，那么dp[2] = 1。令i从3遍历到n，检查A[i]-A[i-1]和A[i-1]-A[i-2]是否相等，如果不相等，那么dp[i]不改变还是0，因为第A[I]和A[I-1]、A[I-2]连续的3个数不能构成等差数列。如果相等，那么dp[i] = dp[i--1]+1，之前以i-1为结尾的数列再加上A[i]又能构成等差数列，所以dp[i]比dp[i-1]多1。最后将dp的所有数字相加就能得到结果。

这道题与上面那一题的差别是数列可以不连续，也就是每个数字可以和前面不同的数字组成等差数列，那就会有不同的差，所以需要在上一题的基础上给dp增加一维记录差。dp[i][diff]表示以A[i]结尾，与前面数字差别为diff的数列个数。i从0遍历到n，j从0遍历到i，计算每一个A[i]-A[j]的值为diff，那么以i为结尾，diff为差的数列个数就加1，如果dp[j][diff]不为0，说明以j为结尾，diff为差的数列存在，这时候就至少有3个数可以组成等差数列（i，j，j前面与j差为diff的数），以j为结尾，diff为差的每一个等差数列再加上A[i]之后，都可以组成新的等差数列，所以dp[i][diff] = dp[i][diff] + dp[j][diff]。最后将所有的dp[i][diff]相加就得到结果了。特别注意这题整数的范围是 -231 到 231-1，所以要用long型。

class Solution {public:    int numberOfArithmeticSlices(vector<int>& A) {        int n = A.size();        vector<unordered_map<int,int> > dp(n);//长度为n的vector，每一个值是不同的diff对应的数列个数        int res = 0;        for(int i = 0; i < n; ++i)        {            for(int j = 0; j < i; ++j)            {                long diff = (long)A[i] - (long)A[j];                if(diff > INT_MAX || diff < INT_MIN)//判断是否越界                    continue;                diff = (int)diff;                if(!dp[i].count(diff))//空处理                    dp[i][diff] = 0;                dp[i][diff]++;                if(dp[j].count(diff) != 0)//前面有数字能组成等差数列                {                    res += dp[j][diff];//累加结果                    dp[i][diff] += dp[j][diff];//状态转移                }            }        }        return res;    }};