LeetCode解题报告（297)-- 413. Arithmetic Slices

Problem

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.

For example, these are arithmetic sequences:

1
2
3

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic.

1	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 the 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:

1
2
3

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.

Analysis

题目给出一个数组，要求我们计算出其中有多少个等差数列切片。其中等差数列切片的定义是长度至少为3的等差数列。遇上等差数列，而且求的是数量，优先考虑DP。我们定义dp[i]是以i结尾的等差数列中有多少个数列切片。状态转移非常简单，只需要在前一个状态的基础上+1即可。

Solution

无

Code

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

Summary

这道题目是DP的一道简单应用，在等差数列类型的题目中dp是非常常用的解法。这道题目的分享到这里，谢谢您的支持！