不间断子数组

You are given a 0-indexed integer array nums. A subarray of nums is called continuous if:

• Let i, i + 1, ..., j be the indices in the subarray. Then, for each pair of indices i <= i1, i2 <= j, 0 <= |nums[i1] - nums[i2]| <= 2.

Return the total number of continuous subarrays.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [5,4,2,4]
Output: 8
Explanation: 
Continuous subarray of size 1: [5], [4], [2], [4].
Continuous subarray of size 2: [5,4], [4,2], [2,4].
Continuous subarray of size 3: [4,2,4].
Thereare no subarrys of size 4.
Total continuous subarrays = 4 + 3 + 1 = 8.
It can be shown that there are no more continuous subarrays.

 

Example 2:

Input: nums = [1,2,3]
Output: 6
Explanation: 
Continuous subarray of size 1: [1], [2], [3].
Continuous subarray of size 2: [1,2], [2,3].
Continuous subarray of size 3: [1,2,3].
Total continuous subarrays = 3 + 2 + 1 = 6.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

/*
 * 思路：
 * 使用Java Stream API实现相同的逻辑
 */

import java.util.stream.IntStream;

public class ContinuousSubarraysStream {
    public static int countContinuousSubarrays(int[] nums) {
        return IntStream.range(0, nums.length)
                .mapToObj(i -> new int[]{i, i})
                .mapToInt(p -> {
                    int start = p[0];
                    int end = p[1];
                    while (end < nums.length && (end == start || Math.abs(nums[end] - nums[end - 1]) <= 2)) {
                        end++;
                    }
                    return end - start;
                })
                .sum();
    }
}

该题解采用了一个双指针技术来确定符合条件的不间断子数组。指针i作为子数组的起始点，而指针j从i开始向后遍历，检查每个新元素是否能扩展当前的不间断子数组。为了判断是否满足不间断条件，算法维护了当前子数组中的最大值maxV和最小值minV，并更新这两个值随着j的增加。如果这两个值的差大于2，表明当前子数组已不再满足条件，因此计算到目前为止的所有可能子数组数目，并调整i的位置来尝试新的子数组。如果j到达数组末尾而没有超过范围，那么将当前子数组的计数加入最终结果中。此外，对于单个元素的子数组，最后结果需再加上n。

O(n^2)

O(1)