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- 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
复制代码
- class Solution {
- public List<List<Integer>> minimumAbsDifference(int[] arr) {
- List<List<Integer>> res = new ArrayList<List<Integer>>();
- Arrays.sort(arr);
-
- int min = arr[1] - arr[0];
- int len = arr.length;
- for(int i =0 ; i <len - 1 ; i++){
-
-
- if( arr[i+1] -arr[i] < min) min = arr[i+1] -arr[i];
-
- }
-
- for(int i =0 ; i <len - 1 ; i++){
-
- List<Integer> list = new ArrayList<>();
-
-
- if( arr[i+1] -arr[i] == min) {
-
- list.add(arr[i]);
- list.add(arr[i+1]);
-
-
- }
- if(list.size() == 2){
- res.add(list);
- list = new ArrayList<>();
- }
-
-
- if(list.size() == 2){
- res.add(list);
- list = new ArrayList<>();
-
- }
-
- }
-
- return res;
- }
- }
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