leetcode-1200-easy

Minimum Absolute Difference

Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements.

Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows

a, b are from arr
a < b
b - a equals to the minimum absolute difference of any two elements in arr
Example 1:

Input: arr = [4,2,1,3]
Output: [[1,2],[2,3],[3,4]]
Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
Example 2:

Input: arr = [1,3,6,10,15]
Output: [[1,3]]
Example 3:

Input: arr = [3,8,-10,23,19,-4,-14,27]
Output: [[-14,-10],[19,23],[23,27]]
Constraints:

2 <= arr.length <= 105
-106 <= arr[i] <= 106

思路一：数组排序后，对数组进行遍历，对于相邻的两个数字，完全分类，只可能出现三种情况，对这三种情况分别进行处理即可

• diff 值小于最小
• diff 值相等
• diff 值大于最小

public List<List<Integer>> minimumAbsDifference(int[] arr) {
    Arrays.sort(arr);

    List<List<Integer>> result = new ArrayList<>();
    int min = Integer.MAX_VALUE;
    for (int i = 0; i < arr.length - 1; i++) {
        int diff = arr[i + 1] - arr[i];
        if (diff < min) {
            result.clear();

            List<Integer> list = new ArrayList<>();
            list.add(arr[i]);
            list.add(arr[i + 1]);
            result.add(list);

            min = diff;
        } else if (diff == min) {
            List<Integer> list = new ArrayList<>();
            list.add(arr[i]);
            list.add(arr[i + 1]);
            result.add(list);
        }
    }

    return result;
}