Leetcode[413] Arithmetic Slices

_ipo 发布于2019-08-16 10:26 / 2138人阅读

摘要：复杂度思路找数组里面的等差数列的个数。想法是如果一开始三个数就满足是等差数列的话，就在当前已有的数目上不断的累加的结果。

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.

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

复杂度
O(N)

思路
找数组里面的等差数列的个数。想法是如果一开始三个数就满足是等差数列的话，就在当前已有的数目上不断的累加count的结果。然后更新sum。

代码

public int numberOfArithmeticSlices(int[] nums) {
    int cur = 0, sum = 0;
    for(int i = 2; i < nums.length; i ++) {
        if(nums[i] == nums[i - 1] && nums[i - 1] == nums[i - 1]) {
            cur += 1;
            sum += cur;
        }
        else {
            cur = 0;
        }
    }
    return sum;
}

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