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  • [LeetCode]413 Arithmetic Slices

    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.




    解题: 如果一个等差数列是长度是n,那么一共有一个 (n-1)(n-2)/2 个子等差数列,所以先求出所有最长的子等差数列的长度即可,利用滑动窗口,两个游标,当当前遇等差数列的条件不满足了以后
    更新计数值和游标,代码如下:
    public int numberOfArithmeticSlices(int[] A) {
            if (A == null || A.length < 3) {
                return 0;
            }  else {
                int count = 0;
                int left = 0, right = 1;
                int diff = A[right] - A[left];
                while (right < A.length) {
                    if (right < A.length - 1 && A[right + 1] - A[right] == diff) {
                        right++;
                    } else {
                        int length = right - left + 1;
                        if (length >= 3) {
                            count += ((1 + (length - 2)) * (length - 2)) / 2;
                        }
                        if (right == A.length - 1) {
                            break;
                        } else {
                            left = right;
                            right++;
                            diff = A[right] - A[left];
                        }
                    }
    
                }
                return count;
            }
        }



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  • 原文地址:https://www.cnblogs.com/javanerd/p/6058789.html
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