[LeetCode] 2210. Count Hills and Valleys in an Array

You are given a 0-indexed integer array nums. An index i is part of a hill in nums if the closest non-equal neighbors of i are smaller than nums[i]. Similarly, an index i is part of a valley in nums if the closest non-equal neighbors of i are larger than nums[i]. Adjacent indices i and j are part of the same hill or valley if nums[i] == nums[j].

Note that for an index to be part of a hill or valley, it must have a non-equal neighbor on both the left and right of the index.

Return the number of hills and valleys in nums.

Example 1:
Input: nums = [2,4,1,1,6,5]
Output: 3
Explanation:
At index 0: There is no non-equal neighbor of 2 on the left, so index 0 is neither a hill nor a valley.
At index 1: The closest non-equal neighbors of 4 are 2 and 1. Since 4 > 2 and 4 > 1, index 1 is a hill.
At index 2: The closest non-equal neighbors of 1 are 4 and 6. Since 1 < 4 and 1 < 6, index 2 is a valley.
At index 3: The closest non-equal neighbors of 1 are 4 and 6. Since 1 < 4 and 1 < 6, index 3 is a valley, but note that it is part of the same valley as index 2.
At index 4: The closest non-equal neighbors of 6 are 1 and 5. Since 6 > 1 and 6 > 5, index 4 is a hill.
At index 5: There is no non-equal neighbor of 5 on the right, so index 5 is neither a hill nor a valley.
There are 3 hills and valleys so we return 3.

Example 2:
Input: nums = [6,6,5,5,4,1]
Output: 0
Explanation:
At index 0: There is no non-equal neighbor of 6 on the left, so index 0 is neither a hill nor a valley.
At index 1: There is no non-equal neighbor of 6 on the left, so index 1 is neither a hill nor a valley.
At index 2: The closest non-equal neighbors of 5 are 6 and 4. Since 5 < 6 and 5 > 4, index 2 is neither a hill nor a valley.
At index 3: The closest non-equal neighbors of 5 are 6 and 4. Since 5 < 6 and 5 > 4, index 3 is neither a hill nor a valley.
At index 4: The closest non-equal neighbors of 4 are 5 and 1. Since 4 < 5 and 4 > 1, index 4 is neither a hill nor a valley.
At index 5: There is no non-equal neighbor of 1 on the right, so index 5 is neither a hill nor a valley.
There are 0 hills and valleys so we return 0.

Constraints:
3 <= nums.length <= 100
1 <= nums[i] <= 100

统计数组中峰和谷的数量。

给你一个下标从 0 开始的整数数组 nums 。如果两侧距 i 最近的不相等邻居的值均小于 nums[i] ，则下标 i 是 nums 中，某个峰的一部分。类似地，如果两侧距 i 最近的不相等邻居的值均大于 nums[i] ，则下标 i 是 nums 中某个谷的一部分。对于相邻下标 i 和 j ，如果 nums[i] == nums[j] ， 则认为这两下标属于 同一个 峰或谷。

注意，要使某个下标所做峰或谷的一部分，那么它左右两侧必须 都 存在不相等邻居。

返回 nums 中峰和谷的数量。

思路一 - 去掉重复元素

类似26题那样，先去掉重复元素，然后再统计峰和谷。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public int countHillValley(int[] nums) {
        int n = nums.length;
        int j = 1;
        for (int i = 1; i < n; i++) {
            if (nums[i] != nums[j - 1]) {
                nums[j] = nums[i];
                j++;
            }
        }

        int count = 0;
        for (int i = 1; i < j - 1; i++) {
            if (nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) {
                count++;
            }
            if (nums[i] < nums[i - 1] && nums[i] < nums[i + 1]) {
                count++;
            }
        }
        return count;
    }
}
// similar to 26

思路二 - 状态机

还是判断每两个相邻元素的大小关系，并用一个变量记录上一对元素比较的结果。如果当前两个元素的比较结果和上一个比较结果不同，则说明当前元素是峰或谷。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public int countHillValley(int[] nums) {
        int res = 0;
        int prestate = 0;
        int n = nums.length;
        for (int i = 0; i < n - 1; i++) {
            int cur = nums[i];
            int next = nums[i + 1];
            if (cur > next) {
                if (prestate == 1) {
                    res++;
                }
                prestate = -1;
            } else if (cur < next) {
                if (prestate == -1) {
                    res++;
                }
                prestate = 1;
            }
        }
        return res;
    }
}