leetcode 413. Arithmetic Slices 等差子数组 + 一个很简单DP动态规划做法

绝地灬酷狼 2022-06-04 08:14 137阅读 0赞

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, 9  
7, 7, 7, 7  
3, -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.

题意很简单，我的最直接的方法就是直接计算difference，然后做统计，然后计算即可，但是网上看到了一个DP动态规划的做法，很棒的做法，dp\{i\}表示以i结束的序列的合理的数量，然后遍历求解即可。

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    class Solution
    {
    public:
        int numberOfArithmeticSlices(vector<int>& a)
        {
            if (a.size() <= 2)
                return 0;
    
            //dp[i]表示以i结束的ArithmeticSlices的数量
            vector<int> dp(a.size(), 0);
            for (int i = 2; i < a.size(); i++)
            {
                if (a[i] - a[i - 1] == a[i - 1] - a[i - 2])
                    dp[i] = dp[i - 1] + 1;
            }
            return accumulate(dp.begin(),dp.end(),0);
        }
    };