leetcode1200. Minimum Absolute Difference

Posted 2022-12-01 seyjs

题目如下：

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

 

解题思路：把arr排好序后依次求相邻两个元素的差值，即可得到结果。

代码如下：

class Solution(object):
    def minimumAbsDifference(self, arr):
        """
        :type arr: List[int]
        :rtype: List[List[int]]
        """
        arr.sort()
        res = []
        min_diff = float(‘inf‘)
        for i in range(len(arr)-1):
            if arr[i+1] - arr[i] < min_diff:
                res = [[arr[i],arr[i+1]]]
                min_diff = arr[i+1] - arr[i]
            elif arr[i+1] - arr[i] == min_diff:
                res.append([arr[i],arr[i+1]])
        return res