HTML要素の高さ、offsetHeight、clientHeight、scrollTopなどについて説明します。

package live.every.day.Programming;

import java.util.ArrayList;
import java.util.Arrays;

/**
 * Given an array nums containing n integers, determine if there are three elements a, b, and c in nums such that a+b+c=0.
 * Find all triples that satisfy the condition and do not duplicate.
 *
 * @author Created by LiveEveryDay
 */
public class ThreeNumbersSumEqualsZero {

    public static ArrayList<ArrayList<Integer>> solve(int[] nums) {

        // triples are represented by ArrayList<Integer>
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();

        if (nums == null || nums.length == 0) {
            return result;
        }

        // Sort
        Arrays.sort(nums);

        for (int i = 0; i < nums.length - 2; i++) {

            // Avoid counting duplicate elements
            if (i ! = 0 && nums[i] == nums[i - 1]) {
                continue;
            }

            /**
             * Fix a number and select two numbers from the ones that follow.
             * Because arrays are ordered, you can use two array subscripts left and right, with left pointing to the last position of the current element and right pointing to the last position of the array.
             * The sum of three numbers is added to the solution set when it equals 0.
             * If the sum of three numbers is less than 0, move left to the right.
             * If the sum of three numbers is greater than 0, move right to the left.
             */
            int left = i + 1;
            int right = nums.length - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0) {
                    ArrayList<Integer> r = new ArrayList<>();
                    r.add(nums[i]);
                    r.add(nums[left]);
                    r.add(nums[right]);
                    result.add(r);
                    left++;
                    right--;

                    // Avoid counting duplicate elements
                    while (left < right && nums[left] == nums[left - 1]) {
                        left++;
                    }

                    // Avoid counting duplicate elements
                    while (left < right && nums[right] == nums[right + 1]) {
                        right --;
                    }
                } else if (sum < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] nums1 = {-1, 0, 1, 2, -1, -4};
        System.out.println(solve(nums1));

        int[] nums2 = {0};
        System.out.println(solve(nums2));

        int[] nums3 = {1, 2, 3, 4, 5, 6};
        System.out.println(solve(nums3));
    }

}

// Output:
/*
[[-1, -1, 2], [-1, 0, 1]]
[]
[]
*/

const height = document.documentElement.offsetHeight
    || document.body.offsetHeight;

const scrollTop = document.documentElement.scrollTop
    || document.body.scrollTop;

// For elements positioned relative to the root node
const top = Element.offsetTop;

// For elements that are not positioned relative to the root node, you need to loop through them
getElementTop(element) {
      let actualTop = element.offsetTop
      let current = element.offsetParent

      while (current ! == null) {
        actualTop += current.offsetTop
        current = current.offsetParent
      }
      return actualTop
}

// There is another way to scroll distance + distance from the top margin of the viewport
const top = Element.getBoundingClientRect().top + window.scrollY.

const top = Element.getBoundingClientRect().top;

const isCSS1Compat = ((document.compatMode || "") === "CSS1Compat");
if (isCSS1Compat) {
    document.documentElement.scrollTop = 100;
} else {
    document.body.scrollTop = 100;
}