[LeetCode] 1200. Minimum Absolute Difference

[grandyang]

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

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 需要清空。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;
    }
};

Github 同步地址:

#1200

参考资料：

https://leetcode.com/problems/minimum-absolute-difference/

https://leetcode.com/problems/minimum-absolute-difference/discuss/388289/Java-sorting-beats-100-explained

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