[LeetCode] 3101. Count Alternating Subarrays

You are given a binary array nums.

We call a subarray alternating if no two adjacent elements in the subarray have the same value.

Return the number of alternating subarrays in nums.

Example 1:
Input: nums = [0,1,1,1]
Output: 5

Explanation:
The following subarrays are alternating: [0], [1], [1], [1], and [0,1].

Example 2:
Input: nums = [1,0,1,0]
Output: 10

Explanation:
Every subarray of the array is alternating. There are 10 possible subarrays that we can choose.

Constraints:
1 <= nums.length <= 105
nums[i] is either 0 or 1.

交替子数组计数。

给你一个二进制数组nums 。

如果一个子数组中 不存在 两个 相邻 元素的值 相同 的情况，我们称这样的子数组为 交替子数组 。

返回数组 nums 中交替子数组的数量。

思路

这道题不难。通过这道题我们可以学到如何计算交替数组的长度。这个思路有助于我们解决一些其他题目，详见相关题目。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public long countAlternatingSubarrays(int[] nums) {
        int n = nums.length;
        long res = 0;
        long count = 0;
        for (int i = 0; i < n; i++) {
            if (i > 0 && nums[i] != nums[i - 1]) {
                count++;
            } else {
                count = 1;
            }
            res += count;
        }
        return res;
    }
}

相关题目

3101. Count Alternating Subarrays
3206. Alternating Groups I
3208. Alternating Groups II