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;
}
};
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