目录


题目地址:https://leetcode.com/problems/minimum-absolute-difference/

题目描述

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:


  1. ​2 <= arr.length <= 10^5​
  2. ​-10^6 <= arr[i] <= 10^6​

题目大意

给出了一个由不同数字构成的数组,哪些数字的差等于所有数字之差的最小值。

解题方法

排序

这个题肯定要先求所有数字差的最小值,暴力算是O(N^2)不可取。我们知道两个数字差最小,肯定是这两个数字比较接近,所以我们可以先排序,然后找到相邻数字的差的最小值。

找出所有数字差的最小值之后,再遍历一次,找出哪些相邻的数字差等于该最小值就行了。

C++代码如下:

class Solution {
public:
vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
const int N = arr.size();
sort(arr.begin(), arr.end());
int min_diff = INT_MAX;
for (int i = 0; i < N - 1; ++i) {
min_diff = min(min_diff, arr[i + 1] - arr[i]);
}
vector<vector<int>> res;
for (int i = 0; i < N - 1; ++i) {
if (arr[i + 1] - arr[i] == min_diff) {
res.push_back({arr[i], arr[i + 1]});
}
}
return res;
}
};

日期

2019 年 9 月 22 日 —— 熬夜废掉半条命