1200. Minimum Absolute Difference

QUESTION:

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 <= 10^5 -10^6 <= arr[i] <= 10^6

EXPLANATION:

emmm,也是一道easy的题目,思路比较简单,1,首先需要找到最小距离。2.遍历将符合最小距离的结果放在集合中。
基本上明白了这个思路也就可以直接写出代码了。

SOLUTION:

class Solution {
    public List<List<Integer>> minimumAbsDifference(int[] arr) {
        Arrays.sort(arr);
        int min = Integer.MAX_VALUE;
        for(int i = 1;i<arr.length;i++)
            min = Math.min(min,arr[i]-arr[i-1]);

        ArrayList<List<Integer>> list = new ArrayList<>();
        for(int i = 1;i<arr.length;i++){
            if(arr[i]-arr[i-1]==min){
                ArrayList<Integer> tmp = new ArrayList<>();
                tmp.add(arr[i-1]);
                tmp.add(arr[i]);
                list.add(tmp);
            }
        }
        return list;
    }
}
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