[Leetcode] 413. Arithmetic Slices 解题报告

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.

思路：

1、动态规划：定义dp[i]表示以A[i]结尾的等差数列的个数，则状态转移方程为：如果A[i]加入之后不破坏以A[i-1]结尾的等差数列，则dp[i] = dp[i-1] + 1；否则dp[i] = 0。该算法的时间复杂度是O(n)，空间复杂度也是O(n)。

2、找规律：这个方法本质上还是动态规划，只是把空间复杂度进一步降低到了O(1)，确实还挺巧妙的，详细解法请见：http://blog.csdn.net/camellhf/article/details/52824234。

代码：

1、动态规划:

class Solution {public:    int numberOfArithmeticSlices(vector<int>& A) {        int n = A.size();        if (n < 3) {            return 0;        }        vector<int> dp(n, 0);   // dp[i] means the number of arithmetic slices ending with A[i]        if (A[2] - A[1] == A[1] - A[0]) {            dp[2] = 1;          // if the first three numbers are arithmetic or not        }        int result = dp[2];        for (int i = 3; i < n; ++i) {            if (A[i]-A[i-1] == A[i-1]-A[i-2]) {                dp[i] = dp[i-1] + 1;            }            result += dp[i];    // accumulate all valid slices        }        return result;    }};

2、找规律：

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