题目大意:返回所有子数组中等差数列的个数。
Problem
https://leetcode.com/problems/arithmetic-slices/description/
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.
Solution 0: Reduction
Reduce the problem to # of all 1 sub arrays.
B[i – 2] = is_slice(A[i], A[i+1], A[i+2])
Time Complexity: O(n)
Space Complexity: O(n)
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// Author: Huahua // Running time: 4 ms class Solution { public: int numberOfArithmeticSlices(vector<int>& A) { if (A.size() < 3) return 0; vector<int> B(A.size() - 2, 0); for (int i = 2; i < A.size() ; ++i) { if (A[i - 1] - A[i - 2] == A[i] - A[i - 1]) B[i - 2] = 1; } return all1SubArrays(B); } private: // return number of arrays whose values are all ones. int all1SubArrays(const vector<int>& A) { int sum = 0; int curr = 0; for (int i = 0; i < A.size(); ++i) if (A[i]) sum += ++curr; else curr = 0; return sum; } }; |
Solution 1: Combined
C++
Time complexity: O(n)
Space complexity: O(1)
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// Author: Huahua // Running time: 4 ms class Solution { public: int numberOfArithmeticSlices(vector<int>& A) { int sum = 0; int curr = 0; for (int i = 2; i < A.size() ; ++i) { if (A[i - 1] - A[i - 2] == A[i] - A[i - 1]) { sum += ++curr; } else { curr = 0; } } return sum; } }; |
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