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Description
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, bare fromarra < bb - aequals to the minimum absolute difference of any two elements inarr
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
这道题给了一个没有重复数字的整型数组，现在让找出差的绝对值最小的数对儿。既然是 Easy 的身价，那么就没有太 Fancy 的解法，为了更方便的找出差的绝对值最小的数对儿，先给数组进行排序，这样最小差值一定会出现在相邻的两个数字之间。接下来就是遍历所有相邻的两个数字，维护一个最小值 mn，若当前差值 diff 小于等于 mn，则进行进一步操作，二者中唯一不同的是当 diff 小于 mn 时，结果 res 需��清空。然后将 mn 更新为 diff，并把当前数组对儿加入到结果 res 中即可，参见代码如下：
class Solution {
public:
vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
vector<vector<int>> res;
int n = arr.size(), mn = INT_MAX;
sort(arr.begin(), arr.end());
for (int i = 1; i < n; ++i) {
int diff = arr[i] - arr[i - 1];
if (diff <= mn) {
if (diff < mn) res.clear();
mn = diff;
res.push_back({arr[i - 1], arr[i]});
}
}
return res;
}
};
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